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MODERN ALGEBRA 
THIRD SEMESTER COURSE 


BY 

WEBSTER WELLS, S.B. 

» • 

AUTHOR OF A SERIES OF TEXTS ON MATHEMATICS 


AND O' 

WALTER W.MIART, A.B. 


ASSOCIATE PROFESSOR OF MATHEMATICS, SCHOOL OF EDUCATION 

UNIVERSITY OF WINSCONSIN 

AND TEACHER OF MATHEMATICS, WISCONSIN HIGH SCHOOL 




) 


D. C. HEATH AND COMPANY 

BOSTON NEW YORK CHICAGO 

ATLANTA SAN FRANCISCO DALLAS 

LONDON 




This book may he had with or without answers 
at the same 'price. Answer books, bound in 
paper, may be obtained free of charge by teachers. 


COPYRIGHT, 1929, 

BY EMILY R. WELLS AND WALTER W. HART 
2 I 9 


PRINTED IN U.8.A. 

©CIA 14100 

OCT 11 I9?9 



PREFACE 


This text supplies subject matter and instruction for the 
third semester course in algebra. A clearly indicated basic or 
minimum course is offered for those schools which require or 
desire a brief course; equally clearly indicated optional topics 
extend this basic course for schools which can do more than a 
minimum course. 

Special efforts have been made to prevent the first seven 
chapters from becoming a mere “re-do” of the first course in 
algebra. Diagnostic Tests covering subject matter about which 
the pupils know something from their first course appear in 
these chapters to help both the pupils and the teacher discover 
what parts of these chapters need special attention. They can 
serve not only as a means of diagnosis but as a challenge to the 
pupils; they are designed to stimulate the pupils to a desire 
for self-diagnosis, and to a sense of responsibility for self- 
improvement. Of course, any teacher who does not wish to 
use them for these purposes can omit them altogether or can 
use them as additional practice material or as pre-tests toward 
the end of the chapter in preparation for subsequent tests to 
be given by the teacher. 

The Diagonostic Tests are followed by remedial instruction 
and practice of a relatively simple sort. These parts of the 
seven chapters have been extended a bit more than in preceding 
texts for this grade by the same authors to compensate for the 
simplification that has taken place in the first course in algebra 
in recent years. Pupils and schools not needing this elementary 
practice should omit it by all means. For such there are in¬ 
cluded more stimulating examples in lists having the suffix 


IV 


PREFACE 


“b.” (See Ex. 4, b, p. 10.) Besides, there are in these chapters 
clearly marked new topics (See pp. 21, 32, 39, 88, 110) some of 
which are a part of the minimum course, and some of which are 
optional. (See pp. 21, 23, 28, 52, 57, etc.) By these means these 
chapters are placed on a higher plane than corresponding chap¬ 
ters in the first course. 

Attention is called to the chapter on Functional Relationship. 
(See p. 59.) The desire to place in the hands of teachers and 
pupils a satisfactory treatment of this subject, which has come 
to be stressed in recent years, was one of the chief reasons for 
writing this new text. The treatment will be found simple and 
adequate without being verbose or extended unnecessarily. 

The basic or minimum course includes, first, as much of the 
remedial instruction and practice as the class needs; second, 
those topics which are not marked optional’^; and third, those 
exercises and problems which are not accompanied by the suffix 
‘‘b” or which do not accompany an optional topic. This basic 
course includes all that most colleges can possibly require from 
candidates for admission. Whether or not a particular school 
or individual pupils should be expected to master this minimum 
course must be determined by local conditions. 

The optional topics and more difficult examples are included 
as part of the long established policy of Wells and Hart texts 
of providing materials for the able pupils and classes beyond 
the mere basic requirements. To neglect such pupils is just as 
great a fault educationally as it is to overtax the less capable 
ones. Only by some such means can the full fruits of segrega¬ 
tion of pupils into ‘‘X, Y, and Z” sections be secured and the 
widely discussed and generally admitted needs of pupils having 
“differences in ability” be provided for. In a Handbook for 
Teachers which accompanies this text, the author offers sug¬ 
gestions about means of utilizing such optional material, gath¬ 
ered from his own extensive experience. 


PREFACE 


V 


Chapter VII gives only the subject of square root and quad¬ 
ratic surds, as in forerunners of this text. The subject of cube 
root, and other radicals, and the formal operations with them 
and exponents have no significance as preparation for the chap¬ 
ter on quadratic equations, and little significance outside of 
college entrance requirements. Such parts of radicals and ex¬ 
ponents as are required for admission to certain colleges appear 
in Chapter XII; this chapter can be taught immediately after 
Chapter VII if the teacher wishes. 

Chapter VIII includes a complete treatment of Quadratic 
Equations. This chapter presents an innovation in that the 
Theory of Quadratics is included in it instead of being left for 
a remote chapter. In particular, part of this ^‘theory” is made 
to function in the course by being used as a means of checking 
the solution of a quadratic. (See foot of pp. 120, 123, and 
p. 127.) Attention is directed also to the means of motivating 
this whole chapter which appears on pages 114 and 115. 

Graphs do not appear as a separate chapter in this text. They 
are made to function as a vital part of the teaching procedure, 
being used as a most valuable and vivid means of illuminating 
otherwise abstract subject matter. (See pp. 60, 65, 76, 78, 115, 
116, 118, etc.) So used, graphs are an important part of really 
fused mathematics — not mere general mathematics. 

Attention has been given to modern views about accuracy in 
computation. (See p. 68.) In the chapters on trigonometry arid 
logarithms these views are used, as well as in the computations 
introduced under the subject of the formula. 

The chapter on trigonometry gives complete instruction in 
that part of the subject which is now a part of the first course 
in algebra, and, besides, it includes the solution by logarithms 
of the same kind of problems, for this is now a part of the 
requirements in third semester algebra for certain institutions. 

As parts of a systematic teaching procedure, this text pro- 


VI 


PREFACE 


vides full instruction on each topic, an adequate amount of 
practice exercises, additional exercises at the back of the text, 
cumulative reviews, and chapter mastery-tests. (See for these 
last the close of each of the chapters after the first few.) In 
so doing this text recognizes the progress made in the theory 
of teaching and furnishes teachers the means of applying this 
valuable theory in this particular subject. 

In style and workmanship this text will be found a fit sequel to 
its predecessor, the Wells and Hart Revised Modern First Year 
Algebra. 


I 


CONTENTS 


PAGE 

Preface . iii 

CHAPTER 

I. The Fundamental Operations. 1 

II. Special Products and Factoring. 15 

III. Fractions. 29 

IV. First Degree Equations. 45 

V. Functional Relationship. 59 

VI. Systems of First Degree Equations . 75 

VII. Square Root and Quadratic Surds. 99 

VIII. Quadratic Functions and Equations .114 

IX. Graphs of Equations of Second Degree; Two 

Variables.'.145 

X. Systems Involving Quadratics .151 

XI. Factors and Equations of Higher Degree; Fac¬ 
tor Theorem. 164 

XII. Exponents and Radicals.170 

XIII. Logarithms.184 

XIV. Progressions.196 

XV. The Binomial Theorem.214 

XVI. Trigonometry.221 

XVII. Variation.231 

Additional Exercises.237 

Index.250 

Table of Squares, Cubes, and Roots.253 

Table of Sines, Cosines, and Tangents .... 257 

Table of Logarithms of Sines, Cosines, and 

Tangents.262 

vii 
























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I 






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ALGEBRA 


THIRD SEMESTER 


I. THE FUNDAMENTAL OPERATIONS 


1 . How much elementary algebra do you still know? Can 
you make 100% on the following test? 


DIAGNOSTIC TEST 1 


Signed Numbers and Algebraic Expressions 


How much is: 

1. (+ 5) + (+ 3)? 

2. (+ 9) + (- 3)? 

3. (- 10) + (+ 4)? 

4. (- 6) + (- 4)? 
6. (+ 6) - (+ 3)? 

6. (- 7) - (- 2)? 

7. (+ 8) - (- 4)? 

8. (- 5) - (+ 7)? 

9. (+ 7) X (+ 3)? 

10. (- 5) X (- 4)? 

21 . 3 x^y is called a _ 

22. a + b is called a _ 


11. (+ 6) X (- 2)? 

12. (- 7) X (+ 4)? 

13. (+ 12) H- (+ 2)? 

14. (- 9) - (- 3)? 

15. (+ 6) 4- (- 2)? 

16. (- 10) - (+ 2)? 

17. 52 = ? 

18. 43 = ? 

19. 24 = ? 

20 . (- 2)3 = ? 


23. a: + 2/ — z is called a_ 

24. X, y, and z are called_ oi x + y — z. 

25. In 2 :x?y^, 2 is the_; 3 is the_of x, 

and 4 is the_of y. 

26. The numerical value of 3 x depends upon the _ 

__When X increases, then _ 


Note. Now read paragraphs 2 and 3 to learn how to answer any 
questions which you missed. If you made less than 100%, do all of 
Remedial Practice 1. 


1 














2 


ALGEBRA 


2. Positive and negative numbers are used to designate 
oppositeness. Thus, if + 5 steps means five steps to the right, 
— 5 steps means five steps to the left. 

(а) The absolute value of a positive or a negative number is 
the arithmetical value remaining when the sign of the given 
number is omitted. 

Thus, the absolute value of — 3 or of + 3 is 3. 

(б) The points on a line are often indicated thus: 

-5 -4 -3 -2 -I 0 +1 4-2 ^-3 4-4 -4-5 ■ 

A 

(c) Just as $5 gain combined with $5 loss produces 0, 

so (+ 3) + (— 3) = 0; (+ n) + (— n) = 0. 

(d) Just as $5 gain combined with $3 gain gives S8 gain, 
so (4- 5) + ( + 3) = +8. Similarly ( — 5) + ( — 3) = — 8. 

To add two numbers having the same sign, add their absolute 
values and prefix their common sign. 

(e) Just as $5 gain combined with S3 loss gives $2 gain, 
so (+ 5) + (- 3) = + 2; (- 15) + (+ 5) = - 10. 

To add two numbers having unlike signs, subtract the smaller 
absolute value from the larger and prefix to the result the sign of 
the number having the larger absolute value. 

if) Since (— 3) + (— 7) = — 10, then (— 10) — (— 3) 
= — 7. This result can be secured by adding +3 to — 10; 
that is, (— 10) + (+3) = — 7. This suggests the rule: 

To subtract one number from another, change the sign of the 
subtrahend and add the result to the minuend. 

Thus: (- 6) - (+ 4) = (- 6) + (- 4), or - 10. 

The subtrahend is the number subtracted and the minuend 
is the number from which the subtrahend is subtracted. 

{g) Adding a negative number gives the same result as sub¬ 
tracting the corresponding positive number. 

Thus: (+ 11) + (— 3) gives the same result as (+ 11) — (+ 3). 



THE FUNDAMENTAL OPERATIONS 


3 


{h) For multiplication, the following definition is used: To 
multiply two numbers, multiply their absolute values; make the 
product positive if the numbers have like signs and negative if they 
have unlike signs. 

Thus: (- 6) X (- 7) = + 42; and (- 8) X (+ 3) = - 24. 

{i) To divide one number by another, find the quotient of their 
absolute values; make it positive if they have like signs, and 
fiegative if they have unlike signs. 

Thus: (- 24) -- (+ 6) = - 4; and (- 32) -- (- 16) = + 2. 

The quotient of two numbers means the first divided by the 
second. 

3. An algebraic expression is a number expressed by literal 
and arithmetical numbers. 

The arithmetical value of an algebraic expression depends on 
the numerical value (or values) of the literal number (or num¬ 
bers) in it. 


EXERCISE 1. REMEDIAL PRACTICE 

1. Give the sum, the difference, the product, and the quotient 
of each of the following pairs of numbers: 

a. + 6, and + 2 d. - 12, and - 4 9 ^. + 8, and - 24 

b. - 18, and + 3 e. + 32, and - 4 h. - 7, and + 14 

c. + 15, and — 5 f. — 36, and + 6 i. — 5, and — 3 

2. What is the value of 4^? of 2^? of (- 3)^? of (- 2)^? 

3. What is the value of 3 x^yz^ when x = - 2, y = 3, and 
2 = — 1 ? 

4. What is the value of each of the following expressions for 
the values of x, y, and z given in Example 3? 

a. 5 c. a:2 + y' e. x^ - 2 xyy^ g. xy + yx xz 

b. x?y d. X? -y^ f. z^-bz-S h. x? ^ y^^ 

5. How much is: a. (— 1)^? b. (— 1)®? c. (— 1)^? 


4 


ALGEBRA 


DIAGNOSTIC TEST 2 


Addition and Subtraction of Expressions 
Simplify by adding or subtracting as directed. 


1 . 5a:+3x+7a: 

2 . 1 y - by 

3. 2 — S b 

7. Add: 2x — by + 3 

3 X + 6 2/ — 4 
— 4x — 2y+2 

9. Subtract: b r — b s 
3r + 2s 


4. S ab + 2 ab — 7 ab 
6. 2x-5 + 3x+6 
6. 6/+2-3^-7 
8. Add: 2 x'^ — 3 xy y"^ 

— 3 x^ — 2 xy 

+ xy — 2 

10. Subtract: 3 x^ — 4 — 6 X7y 

2 x 2 + ?/2 


11. Arrange in descending ^powers of a and add: 

2a2 — 3a+a; —4a+3a2 — 6; 7 — b -\- 2 a 


4. A monomial or term consists of numbers connected only by 
signs of multiplication or division; as, 2 x^, or 3 ab. 

5 . If two or more numbers are multiplied together, each of 
them or the product of any number of them is a factor of the 
product. A factor of a product is an exact divisor of the product. 

6. Any factor of a product is called the coefficient of the 
product of the remaining factors. 

Thus: In 2 ah, 2 is the coefficient of ah, 2 a of h, and 2 h of a. 

In a term like 5 x^y, the factor 5 is called the numerical 
coefficient of x^y. A term like x has the numerical coefficient 1. 
A term like —3a has the numerical coefficient — 3, since 
(- 3) X a = - 3 a. 

7. When one number, the base, is used as a factor two or 
more times, the result is a power of the base. An exponent, 
written above and at the right of the base, indicates the num¬ 
ber of times the base is used as a factor. 

Thus: 3^ = 3 • 3 • 3 • 3, or 81; x® = x • x • x; = y ‘ y • y • y • y. 






THE FUNDAMENTAL OPERATIONS 


5 


8. A polynomial consists of two or more terms. 

A binomial is a polynomial having two terms. 

A trinomial is a polynomial having three terms. 

9. Like terms are terms which have the same literal factors. 

Thus: 2x'^y and — are like terms. 

Unlike terms are terms which do not have the same literal 
factors. Thus: 2 x^y and 2 xy‘^ are unlike terms. 

Unlike terms may be like with respect to one or more numbers; 
thus: axy and 2 hxy are like with respect to x and y. 

10. The laws of addition. 

(a) The associative law of addition. The sum of three or 
more numbers is the same in whatever manner the numbers 
are grouped. 

Thus: CL h c = (u-j-b) c = a (b c). 

(b) The commutative law of addition. The sum of two or 
more numbers is the same in whatever manner the numbers 
are arranged. 

Thus: a-{-b-\-c = a-{-c-\-b = c-{-b-\- Cl. . 

11. Addition and subtraction of expressions. 

Rule 1. To add two or more like terms: 

Multiply their common factor by the sum of its coefficients. 

Thus: 2a{x — y) Zb{x — y) - (2a + 35) (a; — y). 

Rule 2. To add polynomials: 

1. Write the polynomials with like terms in vertical columns. 

2. Add the columns of like terms, and connect the results by 
their signs. 

Rule 3. To subtract one term from a like term or one poly- 
nominal from another: 

1 . Write like terms in vertical columns. 

2. Imagine the signs of the terms of the subtrahend changed, 
and add the resulting terms to those of the minuend. 


6 


ALGEBRA 


EXERCISE 2. REMEDIAL PRACTICE 

1 . Arrange in descending powers of x and add: 4 x^ — 

+ 3 a:?/, 7 xi/ — 2 + 6 2/^ and 9 — S x'^ — 8 xy. 

2. What is the numerical coefficient oi 5 n{x — y)? 

3. (a) Are 3 c(x + y) and 4 d{x 4- y) like in any respect? 

{b) What are the coefficients of the common factor? 

4. Add 3(c — d) — 9(a + d), and 4(c — d) + 7(a + b). 

- 6. Subtract 3 5^ — 10 + 5 5 — 2 5^ from 8 — 5 — 4 5^ + 11. 

6. How much more than x^ — 2 x + 3 is 3 x^ + 11 x — 7? 

7. From S — 4 ab — 5 b^ subtract the sum of a? + ab 

— ¥ and 2 4- 4 ab — bb‘^. 

8. What must be added to m? — 8 m — b to give 7 
+ 4 m — 3? 

9. Subtract 9(m + ti) — 3(m — n) from 12 (m — n) 4- 

(m + n). 

10. From 3 — 2 a^y^ + subtract a?^ + a^y^ — 3 y^^. 

11. Add 3 x^” — 2 x”^ + 1 and 5 x^” + 3 x" — 4. 

12. Add ax 4- by 4- cz and Ax — sy 4- tz. 


DIAGNOSTIC TEST 3 

Parentheses Used in Addition and Subtraction 
Remove the parentheses and combine terms. 


1. X + (x + 5) 

2. y 4- {y - 4) 

3. s - (z + 6) 

4. w — {w — 1) 

6. 2x- (-X + 5) 

6. X - y - z) 

7. 3tt+(— a — 5) 


9. a + (6 + (c — d)) 

10. a 4~ {b — (c — d)) 

11. r - (5 + (i - x)) 

12 . r — {s — {t 4- x)) 

13. 2 X - (3 X - (4 X - 1)) 

14. 5 c + (4 - (2 c - 3)) 

15. 2 m — (— m — (m — 1)) 


8. 4 w - (- 2 m + n) 16. 7 c + (- 2 c + (- c + 3)) 


THE FUNDAMENTAL OPERATIONS 


7 


12. Parentheses used in addition and subtraction. 

The expression x -\- {y — z) means that y — z must be added 
to X. Addition of a number is accomplished by writing it with 
its own sign; 

so X {y — z) = X y — z. 

The expression x — (y — z) means that y — z must be sub¬ 
tracted from X. Subtraction of a number is accomplished by 
writing it with its sign changed; 
so X — (y — z) = X — y + z. 

If your teacher wishes, use the following customary rules. 

Rule 1. When removing parentheses preceded by a plus sign, 
do not change the signs of the terms within the parentheses; when 
removing parentheses preceded by a minus sign, change the signs 
of the terms within the parentheses. 

Rule 2. When placing terms within parentheses preceded by 
a plus sign, do not change the signs of the terms; when placing 
terms within parentheses preceded by a minus sign, change the 
signs of the terms. 

Note. Remedial Practice and other examples appear on p. 237. 

DIAGNOSTIC TEST 4 
Multiplication of Algebraic Expressions 

1 . What is the meaning of of of (mn)^? 

2. How may x • x • x • x • x be written more briefly? 

3. Find: (a) y"^ • y^\ (6) x^ • x®; (c) m • m^. 

4. Find: {a) 2 x • 3 y; (6) 4 r • 3 r^; (c) 2 xy • 3 yz. 

6. Find: (a) (- 2 c) • (+ Sa^); (6) (- 4 m^) • (- Zmn). 

6. Multiply 2 a + 3 6 by afe. 

7. Find the product of 2 x — 3 y and 4 x. 

8. Find: {a) 5r(2s-3t); (b) - 3 c(2 c - 5). 

9. Find: (a) x^(x^ — xy + y^); (6) (— 2 x)(3 x^ + x — 1). 

10. Multiply 3x + 2yby2x-3y. 



8 


ALGEBRA 


13. The fundamental laws for multiplication. 

(a) The associative law of multiphcation. The product of 
three or more numbers is the same in whatever manner the 
numbers are grouped. 

Thus: a • 6 • c = {a • h) • c = a • (b ■ c). 

(b) The commutative law of multiplication. The product of 
two’ or more numbers is the same in whatever manner the num¬ 
bers are arranged. 

Thus: a ‘ b ' c = b • a - c = c • b • a. 

(c) The distributive law of multiphcation. If the sum or the 
difference of two or more numbers is multiplied by a number, 
the product may be found by multiplying each of the num¬ 
bers by the multiplier and connecting the results by their signs. 

Thus: a(b + c — d) = ab + ac — ad. 

14. The law of exponents in multiplication. 

Since = x • x • x and x^ = x ' x • x • x, 

then ^'X^ = X'X'X'X'X'X-x, or x'^. 

The exponent of any number in a product equals the sum of its 
exponents in the factors of the product. 

15. Rule. To find the product of two monomials: 

1. Find the product of their numerical coefficients. 

2. Multiply this result by the product of the literal factors, using 
the law of exponents for multiplication. 

Example. (3 x^y){- 4:y^) = - 12 x^y"^^ 

16. Rule. To find the product of a polynomial and a mo¬ 
nomial : 

1 . Multiply each term of the polynomial by the monomial. 

2. Unite the results with their signs. 

This rule is an application of fundamental law (c) of § 13. 

Example. (- 3 a){x y - z) = - 3 aa; - 3 ay + 3 as. 


THE FUNDAMENTAL OPERATIONS 


9 


17. Rule. To find the product of a polynomial and a poly¬ 
nomial : 

1. Multiply the multiplicand hy each term of the multiplier. 

2. Add the partial products. 

It is desirable to arrange both the multiplier and the multipli¬ 
cand in descending or ascending powers of one number. 

Example. Multiply 2 x — 3 y and — 5 — 3 xy. 

Solution, — Sxy — 

_ 2 X — 3 

2 — b x^y — 10 xy^ ^This is 2 x{x^ — Sxy — 5y^) 

— 3 x^y + 9 xy^ -f 15 y^ This is — 3 y{x^ — Sxy—5y^) 
2 x^ — 9 x^y — xy^ -h 15 y^ 

EXERCISE 3. REMEDIAL PRACTICE 

1. Find: a) x^ • x^; h) y^ • y^) c) d) 10^ • 10' 

2. Find: a) (- 2 x)(H- 3 x^); h) {xy){- Syz))c) (- a){- a) 
Multiply: 

3. — 11 by + 5 c^d 11. 

4. -3 xy by — 12 xy^z 12. 

5. 3 by a 13. 

6. — m^n by 5 mn 14. 

7. — xy by — X 15. 

8. 3 — 2 a by 5 a 16. 

9. x^ — xy by 2 xy ^ 17. 

10. x^ — X — 3 by — 1 18. 

Multiply and then collect like terms: 

19. 8 (lx+ 1^) 22. 24 (2-^ + 1) 

+ ^*(9 “3 + 2 ) 

(1 “ i + f) 1^(2 “ “ 3 “ “ 5 “) 


+ 2 mn -\r n^hj m -\- n 
(? — 2 cd + dr hy c + d 
aP — ah — 2Why 2 a — h 
x^ — 3x + 4by2x — 3 
x^ + x— Ibyx — 2 
y3 - 2 y2 + 3 by y - 4 
x^ — xy^ by X + y 
4 + 2 + 1 by 2 — 1 




10 


ALGEBRA 


18. Parentheses used in multiplication. 

Since 5 a: - 2(?/ - 3) = 5 x + [- J(?/ - 3)] 
therefore 

5 X - 2(?/ - 3) = 5 X + (- 2 2 / + 6), or 5 x - 2 z/ + 6. 
We can get this result immediately if we multiply y by — 2, 
and — 3 by —2. 

Example 1. 2 x — 3(x — 2) = 2 x — 3 x + 6, or 6 — x. 

Here x is multiplied by — 3; also — 2 is multiplied by — 3. 
Example 2. Multiply (2 x — 3) and x^ — 3 x — 5. 

Solution. 1. (2 X — 3)(x2 — 3 X — 5) 

2. =2 x(x2 - 3 X - 5) - 3(x2 - 3 x - 5) 

3. = 2 x3 - 6 x2 - 10 X - 3 x2 + 9 X + 15 

4. = 2 x^ — 9 x^ — X + 15. 

Note. In Step 3, — 3x2 jg 3 )^ 2 . _|_ 9^. jg 3 )(— ^x); etc. 


EXERCISE 4, a 

Simplify as in Examples 1 and 2 above. 


1. 6 a - 2(a + 1) 11. 

2. 7 c - 3(2 - c) 12. 

3. 5 m — 4(m + 3) 13. 

4. 8 X — 5(2 + x) 14. 

6. 4(2 c - 1) - 3(c + 1) 15. 

6. 2(5 — x) — 2(3 X — 5) 16. 

7. 6(3 + 20 - (5^ + 11) 17. 

8. 8(2 - s) - 9(1 - 2 s) 18. 

9. - 3(x - 2) + 4(x - 1) 19. 
10. i(4 r - 6) - (9 - 3 r) /20. 


(x + 3)(3 x2 - 2x + 5) 

{m — 2 n) (2 w} + mn — 3 'uF) 
{3 r — s) (F — 4 — s^) 

(2 X - 3 ?/)(x2 - 7 xy - 2y-) 
{a^ — a — 3) (a — 2) 

(2 - a - 3)(a - 2) 

(3c2+ 7c - 9)(5c - 1) 

(x* + 4 xy + 16 y^) (x — 4 y) 
(?7i^ — mx — 6 x^) (5 m — 2 x) 
(2 X — l)(x2 — 4 X — 5) 


EXERCISE 4. b 


1. x-(x + 1) 

2. - y^{y^ - 2 ) 

3. x®(x2“ - x“ + 1) 

4. -3 x"(x2« - 2 X" - 1) 


5. 5 - 3 ) 

6. - 4a^(- 1 + a* - 

7. (x” — 5)(x’^ — 4) 

8. (a* + 60 («^ - hy) 


THE FUNDAMENTAL OPERATIONS 


11 


DIAGNOSTIC TEST 5 
Division of Algebraic Expressions 

1. Find: a. y^’, b. -i- c. dJ 

2. Find: a. 4 2 a:; h. 12 a:®3 c. 10 xy^2 xy 

3. Find: a. — 6 ^ 2 m; h. (+8 ad) 4 a6); 

c. (— 14 c^) -f- (— 2 c) 

4. Divide: 6 a: + 8 ?/ by 2 

6. Divide: 12 — 6 + 4 by 2 x^ 

6. Divide; —4a2 + 8aby — 2a 


7. Find: a. 


6 X - 4 y 


h. 


x^ + X 


x^ — X^ + X 


2 - 1 X 

8. Divide: x^ — 7 x + 12 by x — 4 

9. Divide: 6 x^ + 11 xy — 10 by 2 x + 5 y 
10 . Divide: x^ — 4x^+7x — 6byx — 2 

y 19. In division, the number divided is the dividend; the 
number by which it is divided is the divisor; the result is the 
quotient. If the division is not exact, there is a remainder. 
Always dividend = divisor X quotient + remainder. 

20. Division by zero is impossible. 

For if n 0 = m, then m • 0 = ?^. But m • 0 = 0, and not n. 

21. Division of any number by itself equals 1. 

Thus n n — 1, iov n ■ 1 = n. 

22. The law of exponents in division. 

^ tL' rO t/v 

Since ^ = 
then 

That is, the exponent of any number in a quotient equals its 
exponent in the dividend minus its exponent in the divisor. 

Thus: 7? -7- = x^~^, or x^. 


X • X 

x^ x^ = x^. 


= 1 • 1 • X • X • X, 


Again: 




= 1 • 2 , or 2 . 






12 


ALGEBRA 


23. Rule. To divide a monomial by a monomial: 

1. Find the quotient of their numerical coefficients. 

2. Multiply the result by the product of their literal factors, 
using the law of exponents for division. 

Thus: (- 12 a:V) (2 = - 6 a: 

24. Rule. To divide a polynomial by a monomial: 

1. Divide each term of the polynomial by the monomial. 

2. Unite the results with their signs. 

Thus: (6 - 4 a2 - 2 a) -r- 2 a = 3 o2 - 2 a - 1 

25. Rule. To divide a polynomial by a polynomial: 

1. Arrange the dividend and the divisor in either ascending or 
descending powers of some common letter. 

2. Divide the first term of the dividend by the first term of the 
divisor, and write the result as the first term of the quotient. 

3. Multiply the whole divisor by the first term of the quotient; 
subtract the product from the dividend. 

4. Using the remainder as a new dividend, repeat Steps 1-3. 

Example. Divide 2 + 8 a — + 15 by 2 — 3 a + 5. 

Solution. + o — 1 

2a2 — 3a + 5 I 2a^ — + 8a + 15 

2a^ - 3a3 + 5a^ 

2a3 - 5a2 + 8a + 15 
2a^ — Sa^ + 5a 

- 2a2 + 3a + 15 

— 2a^ + 3a — 5 

Check. “ Dividend = divisor X quotient + remainder 
2o? — 3a + 5 2a^ — a^ + 8a — 5 

a2 + a - 1 +20 

2a^ - 3a3 + 5a2 2a4 - + 8a + 15 

+ 2a^ — 3a2 + 5a Since this is the dividend, 

__ — 2a^ + 3a — 5 the division was performed 

2a^ — a^ + 8a — 5 correctly. 










THE FUNDAMENTAL OPERATIONS 


13 


EXERCISE 5. REMEDIAL PRACTICE 

Divide and check your solutions: 


6. — 44 a^h + 55 a^h by 11 a?b 

7. — 3 + 6 by — 

8. 21 x^y‘^ — 42 xy^ by — 7 xy"^ 

9. 18 aa:^ — 50 by 2 a 

10. mx^ — 6 mx + 9 m by m 


1. — 10 xy‘^ by + 5 a:?/ 

2. - 24 by - 8 

3. 28 ah^c^ by — 7 

4. 2 — a? -\- ahy a 

5. 6 a:^ — 3 a:^ + 9 a:^ by — 3 a:^ 

11. 15 a;^ — 11 a: — 14 by 3 X + 2 

12. 32 — 15 + 28 xy by 4 X + 5 y 

13. x^ + xy + y^ by x^ + xy + y^ 

14. 4 x^ — y^ + 2 yz — by 2 X — y + z 

15. x^ — 8 y^ by X — 2 y 17. hy c + d 


16. — d'^hy c d 


13. 1 - 
EXERCISE 6 


81 by 1 + 3 m 


Divide: 

1. 4 x^” by — x^ 

2. — 14 by 7 r* 

3. 24 by - 2 aH^ 

4. — 12 x^“y^^ by — 4 x“y^ 


5. 24 a2"*+3 by - 4 

6. — 14 a:2n^3m _ 2 

7. + 2 by 


8. c® 


+ c2 by 


9. 8 y^ — 12y2^ + 16 y^^ by 4 y‘ 

10. 15 x”^ — 10 x” — 5 x'" by — 5 x” 

11. a. Find the sum, the product, and the quotient of 
— 2 x^y^ and 5 x’^y^. 

h. Also of — 3(a — b) and + 4(a — 6). 

12. Simplify x”(x^ - 2) + 2(x- - 1) 

13. Simplify 3(.2 a + .3 6) — 4(.l a — .5 b) 

14. Find (3 X - 5 y)2 - 5 y(x - 5 y) 

15. Find {x + a — b){x — a b) 

16. Divide x^p + 3 x^ - 54 by x^ - 6 


14 


ALGEBRA 


26. The order of performing the fundamental operations. 

In 2 + 3 X 4, if 2 and 3 are added, securing 5, and the result 
is multiplied by 4, the final result is 20. But, mathematicians 
have agreed that in such an expression, the multiplication should 
be done first and afterward the addition; 
so 2 + 3 X 4 = 2 + 12, or 14. 

If the other interpretation were desired, then the expression 
should have been written thus: (2 + 3) X 4. 

By means of parentheses one can always clearly indicate 
which operations are to be done first. The following rule should 
be used when more than one of the fundamental operations is 
to be performed, in case parentheses are not used to indicate 
some other order of performing them. 

Rule. 1, First perform all operations within parentheses or under 
radical signs. 

2. Next do all the multiplications. 

3. Next do all the divisions. 

4. Finally do all the additions and subtractions in the order 
in which they occur. 

Example 1. Find the value of I ^ rt when I = $100, r = .05, 
and f = 4. 

Solution. I -7- rt = 100 -r- .05 X 4 

= 100 .2, or $500 

Most books advise doing the multiplications and the divisions 
in order as they occur. This example would give (100 -i- .05) 
X 4, or 2000 X 4, or $8000. Since the formula used in Exam¬ 
ple 1 gives the principal which produces $100 interest at 5% in 4 
years, this second result is absurd, for $8000 X .05 X 4 = $1600. 

Example 2. If ^ = F -v- find h when V = 770, 
TT = ^Y’ r — 7. 

22 

Solution. /^ = 770 y • 7 • 7; .*. = 770 4- 154, or h = 5. 


IL SPECIAL PRODUCTS AND FACTORING 


27. The following diagnostic test will help you discover what 
you remember about the simple cases in factoring taught in 
the first course in algebra. Make certain that you are skillful 
in these simple cases. 

DIAGNOSTIC TEST 6 
Do mentally and write only the results for: 


1. 

{x + 6)(a: — 6) 

7. 

(2 a 

4" l)(a + 2) 

2. 

{x + 5)(a: + 7) 

8. 

(4 c 

— 3)(c 4- 2) 

3. 

{y - 3)(^ - 9) 

9. 

(x - 

- 5 y)(3 o: 4- 2 y) 

4. 

{2x- b){2x + 5) 

10. 

(f ^ 

- l)(-|o: 4- 1) 

5. 

(m + ll)(m — 4) 

11. 

(3 a 

4~ 2)(3 a 4" 2) 

6. 

(6 - z)(10 + z) 

12. 

(5x 

- 3y)(2o:4- 3 y) 

Factor the following expressions: 




13. 

1 

to 

22. 

2 al 

4“ 5 a 4“ 2 

14. 

+ 6 .r + 9 

23. 

3 62 

- 8 6 4-4 

15. 

- 8 y + 16 

24. 

6c2 

- 11 c 4- 3 

16. 

49 — 

25. 

2 0:2 

4- o: — 6 

17. 

c2 + 5 c + 6 

26. 

CO 

4- 4 y - 4 

18. 

— 7 w + 12 

27. 

10 tt 

|2 — 3 iC — 1 

19. 

— 4 .T — 21 

28. 

7 

— 11 o: — 6 

20 . 

2^ + 6 z — 16 

29. 

12 P 

+ 16^-3 

21. 

+ 4 — 45 F 

30. 

15 r 

* - 4r - 4 


Note. Your score on this test will indicate which parts of pages 16 
to 20 you must study with greatest care. 

15 


16 


ALGEBRA 


28. The product of two binomials of the form ax + h. 

Observe two ways of finding x -\- 2 y) {2 x — b y). 

Solution B. 

(sT-T f^2^x^^ y) 


15 xy 

= ^x^ — \l xy — 10 y’^ 

Such products are to he found always as in Solution B. 

Observe: 1. 6 is the product of 3 x, and 2 x, the first terms. 

2. — 11 is the sum of (+ 4 xy) and (— 15 xy). 

Notice that 4: xy is the product of the two terms which are 

close together; that — 15 xy is the product of the two which are far 
apart. (See the arrows in Solution-B.) 

These products are usually called the cross products because, in 
Solution A, the arrow connecting (2 x) and {+ 2 y) crosses the arrow 
connecting 3 x and (— by). 

3. — 10 ^2 is the product of the second terms. 

Rule. The product of two binonials of the form ax bis the product 
of their first terms, plus the algebraic sum of their cross products, plus the 
product of their second terms. 

Example 1. (2 c - 3 d)(3 c + 5 d) = 6 + cd - 15 

[(- 3 d)(3 c) = - 9 cd) (2 c)(+ 5 d) = + 10 cd; 

(- 9cd) + (+ 10 cd) = + cd.] 

Example 2. x - b){^ x + b) = 0 x^ - 2b 

[The middle term drops out since (- 15 rc) + (+ 15 x) = 0.] 

Example 3. (2 c + 5 d)^ = (2 c + 5 d)(2 c + 5 d) 

= 4 c2 + 20 cd + 25 d2 

[The middle term is twice (2 c)(+ 5 d) since (10 cd) + (10 cd) = 20 cd.] 

Examples 2 and 3 can be solved as in § 29 if your teacher 
wishes. § 29 can be studied before doing Remedial Practice 7. 


T_T 

-\- 4 xy 


Solution A. 

bx^ 4- 4 xy 

— lb xy — 10 y^ 

Q x^ — 11 xy — 10 








SPECIAL PRODUCTS AND FACTORING 


17 


EXERCISE 7. REMEDIAL PRACTICE 

Find the following products mentally: 


1 . {3x + 2){x - 4) 

2. (6 a + 6) (3 a — 2 6) 

3. (9 c — 5 (i)(2 c -f" d) 

4. (5 a — 4 6) (5 a + 4 6) 
6. (n^ — 3 m) + 3 m) 

6 . (r^ — 6 5)(r^ + 2 s) 

7. {x - 2 y^){x 4- 2 1/2) 

8 . (2 a: + 3 y)(2 x - Ay) 

9. (5 a2 + fe2)(3 ^2 + 2 62) 

10. (3 c 4“ d)2 

11 . (4 a - 6)2 

12 . (2 X - 3 y)2 

13. (m — 11 7i)(m 4- 8 n) 

14. (3 x2 — y^)(a:2 + 4 y^) 
/16. (62 - 3 ac) (624-5 ac) 

16. {x - |)(a’ 4- I) 

17. (7a2 - 6)(6a2 4-6) 

18. (Ja - 36)(ia + 36) 

19. (a:2y 4- 7s)(a:2|/ - 12z) 

20. (3 c - 2 d2)2 

21 . {s - nt){s + 3t) 

22. (m2 — 4 ?2)(m2 + 4 n) 

23. (2 a6 — 5)2 

24. {x - i) (x + i) 

25. (3 a: 4- 2)(a: - 11) 

26. (y 4- 10 2)(2/ - Az) 

27. (^5 - At)(lsA- At) 

28. (1-5 A2)(2 + 9 A2) 


29. (2 c — 5 w2)(3 c 4- 5 ^2) 

30. (|m - f)(|m + f) 

31. (4 p — 7 r) (2 p + 3 r) 

32. (12^3 - l){12s^ + 1) 

33. (z2 — 5 m)(z2 4- 11 ■w^) 

34. (5 — 2 y2) (5 ^2) 

35. (7 2/ + i)(7 2/-i) 

36. (4 a - 3 6)2 

37. (4 m— 7 ?i)(2 m 4- 5 w) 

38. (2 — 5 2/z)(4 4- 9 yz) 

39. (x2y 4- 3 a) {x‘^y — 5 a) 

40. (x — 8 c) (x 4- 10 c) 

41. (3 X 4- 5 y) (2 X — 7 y) 

42. (a — 9 t^)(a 4- 8 t^) 

43. (3 c - 2d2)(2c+ 5d2) 

44. (x-f)(x4-f) 

46. (y - 1)2 

46. (6 a 4- 5 6) (3 a — 4 6) 

47. (z - 1)2 

48. (12 X — 9 y)(12 X + 9 y) 

49. (w — f)2 

60. (8 m — 5 w) (3 m + 2 n) 
51. (2 a- i)2 

62. (11 c2 - 8d)(llc2 4- Sd) 

63. (3 X - i)2 

64. (9^4- 4 0(3 5 - 2t) 

65. (f X - f y)(f X 4- f y) 

66. (m - f-)2 


18 


ALGEBRA 


29. The following two special cases of the rule taught in § 28 
save time, and are easily remembered. 

Rule 1. The product of the sum and the difference of the 
same two numbers equals the difference of their squares. 

(x + y){x - y) = - y- 

Rule 2. The square of a binomial equals the square of its first 
term, plus twice the product of its two terms, plus the square of its 
second term. 

{x -\- uY = 2 xy 1/2 

Example 1. 

(3 c - 4 d)2 = (3 c)2 + 2 (3 c)(- 4 d) + (- 4 df 
= 9 c2 - 24 cd + 16 

Example 2. 

[(a: - y) - zjffx - 2 /) + z] = (^r - yfi - 
= x^ — 2 xy + y'^ — 

EXERCISE 8. REMEDIAL PRACTICE 

Find the following products as above. Consider examples 


16 1 

to 24 

optional. 





1. 

(2 a 

+ 5)2 


13. 

m2 4- n){i m2 - n) 

2. 

(3 a 

- 2)2 


14. 

(C2 

- i)(<^ + i) 

3. 

(1 + 4 cf 


15. 

(3 cd + 2)2 

4. 

(x - 

■ 3 yfi 


16. 

C(^ 

+ 

+ 

1—1 
1? 
+ 

I 

6 . 

(ia 

+ 6)2 


17. 

C(r 

+ 5 ) - 2][(r + 5 ) + 2] 

6. 

(c- 

■idy 


18. 

[(a 

+ 3) + 6^[](a + 3) — 6^ 

7. 

(x - 

■ 6 ((T + C 

i<) 

19. 

C(™ 

I 

I 

I—I 

1 

+ 

8. 

(3 a 

— 5 b) {3 a 

+ 5 5) 

20. 

[(a 

+ 5) + 1]2 

9. 

(2 c 

+ 1)(2 c — 

1) 

21. 

[2- 

f ( 2 / + 2 ) J 

10. 

(x^ - 

_ 2/2)2 


22. 

C(c 

- d) + 3]2 

11. 

(2 a? 

' + 3 6)2 


23. 

L(x 

- y) + 2 ? 

12. 

(x^- 

- y^){x^ H- : 

f) 

24. 

C(r 

- s) - tj 


SPECIAL PRODUCTS AND FACTORING 


19 


30. Factoring trinomials of the form ax^ hx -\- c. 

Example. Factor 15 + 17 a: — 4. 

Solution. 1. We seek two binomials whose product is 15x2 + 
17x - 4. 

2. Assume the first terms are 5 x and 3 x; thus; (5 x )(3 x ). 

3. Assume the second terms are 2 and 2, since 2X2 = 4. Arrange 
the signs so that the larger cross product is plus; that is, try 
(5x - 2)(3x + 2). Since the middle term of this product is + 4x^ 
these are not the correct factors. 

4. Assume the second terms are 4 and 1, since 4 X 1 =4. Arrange 
the signs as before, and try (5 x + 4)(3 x — 1). This is not correct for 
the middle term now is + 7 x. 

5. Rearrange and try (5 x — l)(3x + 4). This is correct, for the 
middle term now is + 17 x. 

6. Hence 15 x^ + 17 x — 4 = (5 x — 1) (3 x + 4). 

EXERCISE 9. REMEDIAL PRACTICE 

Factor the following trinomials: 


1. 

+ 10 X + 24 

16. 

2 ■ 

10 

1 

CO 

1 

2. 

2/2 + 11 y + 28 

17. 

3 w? 

+ 4 m. - 7 

3. 

22 - 11 z + 30 

18. 

■ 

1 

1 

4. 

^c2 - 9 w + 18 

19. 

7 - 

- 4c - 11 

6. 

7*2 + 4 r — 21 

20. 

QP +7t-5 

6. 

52 - 2 5 - 15 

21. 

12 f 

- 57/ - 3 

7. 

f + t - 20 

22. 

2z‘ - 

- z - 15 

8. 

m2 — 5 w — 24 

23. 

9P + 6t- 8 

9. 

3 x2 + 5 X + 2 

24. 

15x2 

+ 4x - 3 

10. 

2 — llw + 5 

26. 

5 0* + 16 a6 + 3 62 

11. 

5 m2 + 7 m + 2 

26. 

4 x2 - 

- 24 xy + 35 y‘ 

12. 

6 m2 — 7 m + 2 

27. 

3^2 - 

- 22 + 7 52 

13. 

2 — 0 w 

28. 

4 

+ 4 rs - 15 

14. 

5 c2 - 23 c + 12 

29. 

12 x2 

- 17x + 6 

15. 

8 a2 - 22 a + 15 

30. 

16x2 

— 24 xy + 9 y^ 


Note. Additional examples appear on p. 238. 


20 


ALGEBRA 


31. Two important special cases of factoring. 

Rule 1. The difference of two squares equals the product of the 
sum and the difference of their square roots. 

{x y){x - y) = - y^ 

Rule 1. A perfect square trinomial has two terms which are 
perfect squares, and a third which is the product of their square 
roots. 

2. It has two identical factors. Each consists of the square roots 
of the perfect square terms connected by the sign of the third term. 

Thus; 4:x^ - 12xy + 9 = {2 xY - 2 • {2x){?>y) + {‘^yY 

= {2x - Sy){2x - Sy) 


EXERCISE 10. REMEDIAL PRACTICE 


1. 

2/2 - 64 

21. 

36 a;2 

- 121 

2. 

a;2 - 10 a: + 25 

22. 

100 a2 — 20 a6 + 

3. 

s2 + 6 az + 9 a2 

23. 

9^2 - 

- 12 5^ + 4 f2 

4. 

25 - 1 

24. 

^ w} 

_ 1 

1 6 

6. 

4: c 4c cd -Y d^ 

26. 

1 - 

2 V 

6. 

16 — 8 a; + a:2 

26. 

22 + 


7. 

22;2 — -L ^2 

27. 

a:2 — 

4^14 
-3 X -t Q 

8. 

25 a"^ “h 10 a -\~ 4 

28. 

a;2 — 


9. 

49 62 - 14 5 + 1 

29. 

2/2 + 

y + i 

10. 

9c2-^V 

30. 

22 - 

.25 

11. 

- 12 c2 + 36 

31. 

22?2 — 

.36 

12. 

16 a;2 - 25 

32. 

.49 T 

^ - 1 

13. 

w^ — 16 22? + 64 

33. 

y 2 _ 

.09 22 

14. 

36 - 12 5 + 52 

34. 

22;2 — 

.12 w+ .36 

15. 

c^ -Y d^ — 2 cd 

36. 

a;2 — 

.14 a; + .49 

16. 

2/2 4- 16 2 / + 64 

36. 

.64 + .16 a; + a;2 

17. 

9 — 1 

37. 

.04 T 

^ - 1 

18. 

t f22 _ yi 

38. 

2/2 - 

.09 22 

19. 

1 - 6 6 + 9 62 

39. 

.04 a;' 

^ - .04 a; + 1 

20. 

A - 

40. 

y2 _ 

.06 y + .09 


SPECIAL PRODUCTS AND FACTORING 


21 


Some New Topics in Factoring, Pages 21 to 28 
32. Polynomials factored by grouping (Optional). 
Example 1. Just as aa: + 6a: = Ca + h)x, 
so a(x + 2/) + h(x y) = (a + 6)(a: + y). 

Example 2. Factor 6 a:^ — 15 a:^ — 8 x + 20. 

Solution. 1. 6 — 15 — 8 X + 20 

= (6 x3 - 8 x) - (15 x2 - 20) 

2. =2 x(3 x2 - 4) - 5(3 x^ - 4) 

3. = (2x - 5)(3a« _ 4) 


Factor: 

1. 2 m(x + 2 /) - 3(x + y) 

2. 5 c{r — s) 2 d (r — s) 

3. 4:n(2x - y) - 6(2 x - y) 

4 . 9(p - q) - x(p - q) 

6. a(6 — c) — dQ) — c) 

6. ay -h az -h my + mz 

7 . mr — ms nr — ns 

8. ax — ay — hx + by 

9 . x^ -\- x^ -\r X -\- 1 

10 . 1 /^ + 2 /^ - 3 2 / - 3 

11 . 2 - 5 a - 2 + 5 

12. - 4 z + - 4 

13 . + 3 — 2 m — 6 

14 . 5 ^ + 15 - 5 5 - 3 s^ 

15 . x^y — 2 xy + 4 lX — S 

16 . 2 cd — c^d — 6 + 3 c 

17 . ac + ac^ - 2 6c^ - 2 6c 

18 . ac — 3 be — ad 3 bd 

19 . mr - 3 nr - 3 ns ms 


SE 11 

20 . 2 ax — 4 : ay — bx 2 by 

21 . 5 mr — 3 ns — 3 ms 5 nr 

22 . 2 — 3 — 2 + 3 a 

23. 3 x3 - 6 x2 + X - 2 

24. 2 mx — 3 nx — 2m-\-3n 

25. a^x — a^by + cx — bey 

26. a^b — acd + bed — ab‘^ 

27. 5 a3 - 55 a - 2 + 22 

28. 6 + 3 x^ - 8 X - 4 x^ 

29. 2 ax + 3 6 x - 2 ai/ - 3 6 ?/ 

30. 4 x2 - 2 2 /^ + x?/^ - 8 ?/x 

31. 2 cx + 4 c?x - 3 C 2 / - 6 (^ 2 / 

32. x^ 4 - x^y - xy^ - if 

33 . m^ — m^n — mn^ + n^ 

34. ax - 2 a 2 / + 6 62 / - 3 6 x 

35. 2 X - x^ + 3 x^ - 6 

36. 3 2 /^ + 2 /^ - 2 - 6 2 / 

37. 2 2 /^ — 22 / + 2 2 /^ 2 : — 2 ^ 

38. 2^ - 4sH + sf - 2f 


22 


ALGEBRA 


33. Polynomials reducible to the difference of two squares 

Example 1. Just as = (a: + y)(^x - y) 

I — {m - ny = [a + (m - ?^)][a - (m - rz)] 

= [a 4 - m — nj£a — m + ?i] 

Example 2. Factor a? — + 2 he — 

Solution. 1 . 2 be — — (¥ — 2 he e^) 


3. a2 - 62 + 2 6c 


Factor; 

1. (a - by - c2 

2 . 0^2 _ (2 - yy 

3. m2 — (r + 5)2 

4. (x + zy — ^2 
6. (2 a- by - 9 c2 

6. 16 52 + (^ + ic)2 

7. (x - 2 yy - 4 22 

8. 25 - (r - 3 5)2 

9. (r — 3)2 — 1 

10. 9 - (2 5 + 3)2 

11. a2 + 2 a6 + 62 — c2 

12. m2 — 2 m + 1 — 4 p2 

13. a:2 + 6 a: + 9 - i/2 

14. /2 _ |.2 _ 2 r5 — 52 
16. 4 c2 - a2 + 2 a6 + 62 

Miscellaneous 

31. (a!2 +1)2-4 a;2 

32. (r2 - 3)2 - (r + 1)2 

33. (a: + 1)2 - 5(0: + 1) + 6 

34. (r - 2)2 + 7(r - 2) + 1 


■ c 2 = [a + (6 — c)][a — (6 — c)] 

= [a + 6 — c][o — 6 + c] 

EXERCISE 12 

16. 9 a:2 _ q xy y^ — 

17. 16 a2 + 8 a6 + 62 - c2 

18. 9 r2 — 4 + — 4 5 — 1 

19. ^2 _ 10 / + 25 - 4 a-2 

20. 16 - a:2 + 8 .r - 64 

21. a:2 — 4 a:// + 4 i/2 — 2:2 

22. 62 — 6 6c + 9 c2 — c^2 

23. m2 - 4 7-2 - 8 - 4 52 

24. 9 a:2 — 2.2 — 2 zw — uP' 

25. 9 a2 + 4 6c — 4 c2 — b- 

26. a:2 + 6 a: - 16 i/2 + 9 

27. c2 — 62 — 4 a2 — 4 ct6 

23. A. — 22 xy — 121 x^ — i/2 
29. 16 /v2 - 25 n2 + 1 - 8 k 
39 . 2 AB + A2 - 7^2 + R2 

Factoring Examples 
36. .25 a:2 - 3 a: + 9 

36. 6 a;2 + 5 xi/ — 25 i/2 

37. - A </' 

2 38. + 


SPECIAL PRODUCTS AND FACTORING 


23 


34. Trinomials reducible to the difference of two squares.* 
Type form: + rz/" + 

Example. Factor 64 — 64 aV + 25 

Solution. 1. A 'perfect square containing 64 and 25 is 64 — 

80 a'^m? + 25 m^. It can be obtained from 64 — 64 + 25 by 

subtracting 16 o^m^. Then we must also add 16 to avoid changing 
the value of 64 — 64 + 25 m^. That is, 

2. 64 — 64 + 25 = (64 — 80 + 25 m'^) + 16 a?in^ 

= (8 — 5 + (4 amY 

But this is not factorable, since it is the sum of two squares. 

3. Another perfect square containing 64 a‘^ and 25 w'* is 64 a'* + 80 a'^ni^ 

+ 25 It may be obtained from 64 a'^ — 64 aW + 25 by adding 
to it 144 Then we must also subtract 144 a'^m?. 

That is, 64 — 64 a^m^ + 25 

= (64 a^ + 80 oP’m? + 25 — 144 o^m^ 

4. = (8 a2 + 5 w2)2 - (12 amY 

5. = {(8 + 5 m?) + 12 am } {(8 5 yri^) — 12 am } 

6 . = {8 + 5 m2 + 12 am a? A ^ — 12 am } 


Factor; 

EXERCISE : 

1. a* + aV + h* 

11. 

2. + 1 

12. 

3. x^ + i + 16 

13. 

4. !/^ + 9 1/2 + 81 

14. 

6 . z'* + 6 z‘w‘’ + 25 w 

4 15. 

6 . + 3 rV + 36 s* 

16. 

7. 9 + 2 Tn?fi + <“ 

17. 

8. 16 0 “ - 12 a2 + 1 

18. 

9. 9 ;r* — 10 a;2 -|- 1 

19. 

10. 9 + 2 aV + b* 

20. 

.. + 4 23. 

A 

t. 4 o« + 1 24. 

4 a:^ + 

* This type is not required very 


36 - 16 /2 + 1 
25 — 19 x‘^y'^ + 

4 a 4 + 8 a2 + 9 
9 .T^ + 20 a:2 + 16 
4 + 11 aW + 25 

9 - 34 cH‘^ + 25 

16 - 28 aW + 9 6^ 

9 r4 - 43 rH^ + 49 
16 + 4 a^y + 25 y^ 

4 + 19 m‘^n^ + 49 n'^ 

25. + 64 U 

26. a:^ + 324 


24 


ALGEBRA 


35. To factor an expression is to find two or more expressions 
whose product is the given expression. 

By general agreement the factors must not contain any terms 
which are under a square root sign, or similar radical signs. 

36. A number or an expression is prime if it does not have 
any factors except itself and 1. 

Thus, a, 3, X + y, are all prime numbers. 

37. A monomial factor of an expression is a number which 
will exactly divide each term of the expression; as factor m 
below. 

For: mx + my mz = m{x + y + z) 

Similarly, 

3 x^y + 9 x^y — 3 xy — 3 xy{x^ + 3 .t — 1) 

38. Complete factoring means that the prime factors of an 
expression are to be found. 

Rule. To find the prime factors of an expression: 

1. First remove any monomial factors of the expression. 

2. Then factor the resulting polynomial factor, when possible, 
rewriting all factors which are prime. 

3. Continue until all factors are prime. 

Example 1. Factor 36 ty^ — 36 t^. 

Solution. 1. 36 ty^ — 36 tz^ = 36 t{y^ — z^) 

2 . = 3Qt{y^ + z^){y^ - z^) 

3. = 3Qt{y^ + z^){y^ + z^)(y^ - z^) 

4. = 36 t{y^ + z*){ 2 f + z‘^)iy + z)(y - z) 

Note 1. When you remove a common numerical factor, be certain 

that you remove all of it. In this example, it would not be enough 
to remove 2, 3, 6, 12, or 18. 

Note 2. Remember that y^ + z^ and y^ + 2^ are prime expressions. 

Example 2. Factor 2 ax^ — 12 ax + 18 a. 

Solution. 1. 2 — 12 ox + 18 a = 2 a(x^ — 6 x + 9) 

2. = 2 o(x - 3)(x - 3) 


SPECIAL PRODUCTS AND FACTORING 


25 


EXERCISE 14. a 


Find the prime factors of: 

1. 2 ax — Q ay 9 az 

2. 5 — 10 — 5 y 

3 . Sx^-Sy^ 

4 . am^ — 10 am + 24 a 

5. 12 a^b - 16 ab‘^ 

6 . cy^ — 9 cx‘^ 

7. x"^ - y^ 

8 . — 3 x"^ — X 

9. 9 mx^ — m 

10 . wH -h 4: wt — 60 ^ 

11. - 2a2 - 6 a3 + 8 

12. 4 a^b -25 b 

13 . o!^x — 8 ax — 33 X 

14 . — 5 + 10 X — 5 

16 . x^ — y^ 

16 . 3 — 3 — 60 

17. a^y + 7 aby — 60 b‘^y 

18 . 6 m^x — 7 mx + 2 a: 

EXERCISE 

1. x‘^^ - 7 a:” + 12 

2. - 10 + 24 

3 . + 3 - 40 

4. a:2^ + 2 a:'^ - 24 

6. — 4:m^ — 21. 

6. - 8 a^ - 33 

7. r2y _j. yj/ _ 2 

8. + 2z'^ - 63 

9. x^°' — 6 a:® + 9 


19. x'^b + 8 a:6 — 33 6 

20. a‘^m + 17 am — 38 m 

21. 3 a^ — 108 a 

22. a:“a: — 2 a:s — 6 s 

23. 18 bc^ — 2 bx‘^ 

24. 3 ca:^ + 6 ca: — 9 c 

26. mr^ + 4 mr — 21 r 

26. 18 i + 2 + 2 aH 

27. 24 F - 18 A: - 15 

28. 30 a^b - 120 ab + 120 b 

29. 15 b?x — 10 hx — 15 X 

30. 35 ct"^ + d — 12 c 

31. 15 x^d — 10 x^d — 25 xd 

32. 3c2 + 132 c - 135 
• 33. 45 r% - 80 rV 

34. 3 ¥ - 33 k + 72 
36. — + 25 a: — 100 

36. - a4 + 18 a2 - 77 

, b. {Optional) 

10. z2n _ 2 z^X^ + X^^ 

11. 16 - d^^ 

12. r^c _ fd 

13. x^^ - 25 

14. 3 a2^ + 7 a^ + 2 

16. 6 m‘^y - 11 m^ + 3 

16. 2 x2® - 3 X® - 5 

17. 7r2« - 4r‘^ - 11 

18. 2 y^^ + y^ - 15 


26 


ALGEBRA 


39. Factoring polynomials of the form a” — 6” and a” + b^\ 

By division, as in § 25, it can be proved that: 

(a) -b^ = (a - b){a^ + + b^). 

qB _ b^ — (a — 5)(a^ + a^b + a^b^ + ab^ + b^). 

(b) + 63 = (a + 6 )(a2 - a6 + 6 ^). 

+ 6 ^ = (a + 6 )(a^ — a36 + aW — a¥ + 6 ^). 

(c) a? + 6^, + 6^ a* + 6^, etc., are not exactly divisible 

by a + 6 or a — 6. They are prime number expressions. 

In general we have the following rules. 

40. Rules. Letting n represent a positive integer: 

I. a” + 6”’, when n is odd, is divisible 6 y a + 6. The quotient 

is .6””L 

Observe that the signs of the quotient are alternately plus and 
minus, that of the first term being plus; that the exponent of a starts 
with 1 less than n in the first term, and decreases by 1 in each succeeding 
term; that the exponent of h starts with 1 in the second term, and 
increases by 1 in each succeeding term until it becomes n — 1 in the 
last term. 

II. a” + 6 ”, when n is 2 , 4, 8, 16, etc., is not factorable. When 
n is 6, 9, 12 , etc., a^ + 6” should be factored by Rule I. 

For example: + {iTY 

III. qu _ b^^ when n is odd, is divisible by a — b. The quo¬ 
tient is a^~^ + a^~‘^b + a^~W + a^~'^¥ +.6”“^ 

Observe that the signs are all plus; and that the terms otherwise 
are the same as described in Rule I. 

IV. a” — 6 ”, when n is even, should be factored as the difference 
of two squares. However, it is divisible by a — b, and also by 
a + 6, the quotient being obtained as in Rules III or I respec¬ 
tively. 

The following two special cases are particularly important. 

q3 _ ^3 = ^ qJj ^2) 

= (a + 6)(a2 — a6 + 6^) 




\ 

SPECIAL PRODUCTS AND FACTORING 27 

Example 1. Factor 32 — 243. 

Solution. 1. 32 - 243 = (2 x)^ - 3^ 

2. = (2 X - 3)[(2 x)^ + (2 x)3 • 3 + (2 • 3^ + (2 x) • 3^ + 3^ 

3. = (2 x + 3)(16 + 24 + 36 + 54 X + 81) 

Example 2. Factor x^ + completely. 

Solution. 1. x^ + = (x^)® + {y^Y 

2. = (x^ + y^)l{x^Y - (x3)(^3) + (^3)2] (Rule j) 

3. = (x + y) (x2 - X?/ + 2/2) (x® - x32/3 + 2/®) (See § 22) 

Note 1. x® + 2/® is also divisible by x + 2/ by Rule I. The quotient 

is x* — x^2/ + ~ + ^^y^ ~~ ^y^ + y^- 

Note 2. The correctness of the solution can be checked as usual 
by substituting A^alues for the literal numbers. 

EXERCISE 15 

Factor the following expressions completely, when possible: 
Consider Ex. 33 to Ex. 48 optional. 


1. 

x^ + 

17. 

X3 — 

82/^ 

33. 

x^ — 

16 

2. 

f-S 

18. 

63 + 

27 c3 

34. 

X3 - 

32 

3. 

2/3 + 27 

19. 

C3 + 

8# 

36. 

1 - : 


4. 

vY — 64 

20. 

8x3 

- 1 

36. 

Xl3 - 

1 

5. 

8x3-1 

21. 

8a3 

- 27 63 

37. 

32 a3 

+ 1 

6. 

27 2/'+! 

22. 

27 m 

3 + 

38. 

Xl3 - 

ylO 

7. 

^3 _ 1 

X 3 

23. 

x3 - 

125 y^ 

39. 

— 

V 

8. 

uY + 

24. 

64 + 

■ m® 

40. 

ix3- 

- y^ 

9. 


25. 

x^ — 


41. 

x® — 

64 

10. 

^ + xis 

26. 

a3 - 

64 

42. 

1 ^3 . 
8 

- y^ 

11. 

x® - 8 

27. 

1 - 

64 63 

43. 

2V 2/^ 

— i 

12. 

2/3+27 

28. 

X3 - 

yS 

44. 


- 63 

13. 

x3 + 8 a3 

29. 

X3 + 


46. 

X3 - 

.125 

14. 

23 - 27 

30. 

x3 + 

1 / 

46. 

+ 

.008 

16. 

64 m3 + x3 

31. 

X3 - 

yS 

47. 

x3 — 

.027 

16. 

27 f 

32. 

7^53 - 

- 27 fi 

48. 

Xl2 - 

yl2 


/ 


28 


ALGEBRA 


41. Concerning the factor theorem, and solution of equations 
by factoring. 

Some courses of study require the factor theorem; some make 
it optional. The College Entrance Examination Board does not 
require it. In this text, it appears on page 164. 

Solution of equations by factoring appears on page 120. 

These pages can be studied now if the teacher wishes. 


EXERCISE 16. CHAPTER MASTERY-TEST 


Part a. 

Factor: 

1 . 4:x^ - 121 

2. 2y2 + 11 ^ - 21 

3. 9 — 30 X + 25 

4. 1 - 144 

5. x^ — 2 xy — 99 

6. 12 - 11 s - 5 

7. 16 ^2 - 24 ^ + 9 

8. 15 - 8 2 - 12 


Elementary Cases 

9. 25 x2 - i y2- 

10. x2 - I X -h ^ 

11. 

12. x2 - f X + yV 

13. x2 + 1.6 X + .64 

14. _ 1 21 

15. 14 m2 4- 3 m - 2 

16. x2 - f X + If 


Part h. More Difficult Examples 


1. 

X2n _ 

- 8 x” + 16 

11. 

ax — 

■ 2 ay + bx 

- 2 by 

2. 

9 x2“ 

- 16 

12. 

(x + 

y)2 - 2 X 

- 2y + 1 

3 . 

X2n 

- 2 x^ - 63 

13. 

4 x2 

- y2 + 2 y 

- 1 

4. 

2 x2® 

+ 9 X® — 35 

14. 

3 ax 

— 3 ay — cy -h cx 

5. 

x2-f 

2 xy + y2 — 10 

15. 

+ 

■TT^ 


6. 

f- 

6 y + 9 — 2 ^ 

16. 

x4 - 



7. 

9 - 

a? - 2 ah -b‘^ 

17. 

(x - 

yY-2x 

-f 2y - 15 

8. 

16 - 

x2 + 4 xy — 4 y2 

18. 

C2 - 

A(P + i.d 

- 1 

9. 

8x3 

- 

19. 

am -p an — ^ n 

- \ m 

10. 

i + 

X® 

20. 

x3a _ 

- 



III. FRACTIONS 


DIAGNOSTIC TEST 7 
Can you pass the following test on fractions? 

1. Reduce to lowest terms: (a) (b) ^ 

> 55 d X + 15 

2 . Multiply: (a) g X (b) • 3^^ 

3. Divide: (o) (b) ■ f T ^ 

^ M4 21 ^ ^ - 4 a + 4 a - 

4. Add: («) | + |; W ^ + ^ 

6. Subtract: (a) | (6) ^ ^ 


6. Add: 


X 2 X 


7. Subtract 


X — 4 X + 5 

2 ?/ y 


y - h 2/ + 4 
8. Simplify: - -— 


* — 5 


9 a:-3 x2 + 5x+6 


I. Simplify: (a) (l + g) ^ ~ 

<«(?+>) "(I'-') 


10. If X increases, what happens to - ? 

Note. Mark with a cross (X) the examples you could not solve and 
master the corresponding parts of the following remedial practice. Do 
your best also to learn to solve the new types of examples which appear 
in this chapter. 


29 














30 


ALGEBRA 


Remedial Instruction and Practice, Pages 30 to 31 

42, A fraction indicates the quotient of the numerator 
divided by the denominator. The numerator and denominator 
are called the terms of the fraction. 

43. First fundamental principle of fractions. The numerator 
and denominator of a fraction can be divided by the same number 
without changing the value of the fraction. 

Thus, if both terms of | are divided by 2, we get |, and f = f* 


44. Rule. To reduce a fraction to its lowest terms: 


1 . Find the prime factors of both terms. 

2 . Divide both terms by all their common factors. 


Thus: 


+ 3 g - 28 
g2 — 16 


1 

(a^)(g + 7) 
(g + 4) 

1 


or 


g 7 
g + 4 


Caution. 


^ since x is a factor and can be removed. 
4 :X 4 




cannot be reduced, as x is not a factor. 


X is an 


addend and cannot be removed from both terms without chang¬ 


ing the value of the fraction. 


Thus 


2 + 3 
2 + 4 


does not = 


3 

4‘ 


45. Multiplication and division of fractions. 

Rule. To find the product of two or more fractions: 

1 . Find the prime factors of the numerators and denominators. 

2 . Divide out {cancel) factors common to a numerator and a 
denominator. 

3. Multiply the remaining factors of the numerators for the 
numerator of the product, and of the denominators for the denomina¬ 
tor of the product. 

Rule. To divide one fraction by another: 

1 . Invert the divisor fraction. 

2. Multiply the dividend by the inverted divisor. 







FRACTIONS 


31 


EXERCISE 17. REMEDIAL PRACTICE 


Reduce to lowest terms; 


1 

5 x^ 2 /^ 

R 

a? 

- ¥ 

1« 

3 xy^ 

0. 

{a - 

- by 

o 

12 mV 

n 

3 r 

+12 

z. 

20 m^n? 

0. 

y .2 . 

- 16 

q 

2(a -1- 6) 

7 

- 

- 2/' 

o. 

a2 - ¥ 


X? - 

- 

4 . 

CO 

I 

CO 

0 

4 X 

+ 4 2/ 


3,2 _ ^2 


x4 4- 


^ + m — 56 

im? — m — 42 

10 - 9 ^ + 18 

■ 2/^ + 2/ - 12 

3 — 3 11? 

3 + 6 mn + 3 

cr^ — cr — 12 c 
3 r2 + 13 r + 12 


Perform the indicated operations: 


x2 - a? 

2 X + 2 a 

16. 

xy^ - y^ 

H= 

4 

1 

to 

(x + a)* 

3 X 

a? 4- x^y 

x^ — 2 xy 4- y^ 

4 Ml* - 1 

m 4- 4 

17. 

St 

2t 

m? - 16 

2 m 4 - 1 

2 /^ - 6 2 / + 8 ’ 2 /^ - 2 / - 12 

- d* 

¥4- d‘^ 

1 Q. 

lo 

1 

2 ¥ 4- 2 rs +2 

(c - 

c — d 

lo. 

r — s 

2 r 4- 2 s 

x^ — 1 

— 4 

3 X — 

6 


2x - 4 

x^ — X — 2 

X^ + X ■ 

- 2 


— 1 

4 m? — 

3 n 

m — 1 



a? 4- ¥ S a — Qh a? — 4¥ 

a? + 3 ab + 2 b‘‘ ' 3 a? - Sab + 3 h'^ "" a + 2b 
a?+ 7 ab+ 10 b^ a + b 1 

OO - . - — - 

+ 6 a6 + 5 6^ a? 4 ab 4 ¥ a + 2 6 

a:2 + 3x+9 4-— S — 4 

OO _ . - -J- - 

x‘‘ + X - 12 x? -27 x^- Ox+0 

2 *2 + 7 X - 15 2 X* - 19 X + 42 x^ - x - 30 
2x2-3x-14 2x-3 x+2 













































32 


ALGEBRA 


A New Problem in Fractions, Pages 32 to 35 

46. Three signs of any fraction must be considered, the 
sign of the numerator, of the denominator, and of the fraction 
itself. Certain rules for making changes in these signs follow: 


Rule 1 . If the sign, of either the numerator or the denominator 
of a fraction is changed, the sign of the fraction also must be changed. 

Thus: (a) ^ - L but ±-i = + |. 


In this example, when 
the fraction changes from 
- 4 


4 is changed to + 4, the value of 
J to + ^. 


ih) 


+ 12 


- 3 ; but _ 


3 


In this example, when + 12 is changed to — 12, the value of 
the fraction changes from — J to + ^. 


Similarly, 


+ 5 
- 20 


20 


) or 


.25; - 


+ 10 


+ or .3. 


Example. Reduce to lowest terms 
0 : 2-9 

12 + 2 0: - 2 0:2' 


Solution. 1. The terms of the denominator, not being in the same 
order as those of the numerator, must be rearranged. 


Then 


0:2 — 9 _ 0:2 — 9 

12 + 2 a: - 2 a:2 “ - 2 a:2 + 2 a: + 12 ' 


2. The negative coefficient of is inconvenient. Change the signs 
of the denominator by multiplying by — 1. Then by Rule 1, 

0:2 — 9 _ 0:2 — 9 

- 2 a:2 + 2 a: + 12 ” 2 a:2 - 2 a: - 12 


(a; - 3)(a: + 3) 
2(a:2 - a: - 6) 


1 

— 3) (a; + 3) x + 3 

(x + 2)’ ^ 2{x + 2) ■ 


4. 













FRACTIONS 


33 


Rule 2. If the signs of both the numerator and the de7iominator 
of a fractio7i are changed, the sig7i of the fraction is not changed. 


Thus, 


- 4 
+ 12 


1 

3’ 


and also 


+ 4 
- 12 


1 

3' 


In this example, - 4 is changed to + 4, and + 12 to - 12, 
but the value of the fraction remains — 


Cl - 1 1 — 2 X 2 X 

Similarly,-;;— = — — 

-4y 4:y 


.5 X , 

y ’ 


and + 


— 6 t 


+ 3. 


47. There are two ways to change the signs of an expression 

(such as the numerator or the denominator of a fraction). 

(a) If the expression is not factored, change the sign of each 
term of it by multiplying each of them by — 1. 

Thus: (- 1)(- 2a:2-3a; + 4) = 2a;2 + 3a;-4. 

ib) If the expression is factored, change the signs in an odd 
nu7nber of factors of the expression. 

Thus: 2 • (— 3) • 4 = — 24, but 2 • (+ 3) • 4 = + 24. 

Similarly, consider {x — y) {y — z), 

y — z becomes — y -\- z when its signs are changed. 

Therefore ix — y){y — z) = — {x — 7j){z — y), for changing the 
signs oi y — z changes the sign of the product. 

On the other hand {x — 7j){y — z) = + (y — x){z — y) because the 
signs of both x — y and y — z have been changed. This means that 
— 1 has been used twice as a factor, and therefore the sign of the 
product is not changed. 


Do not confuse “changing the signs of the terms of an ex¬ 
pression” and “rearranging the terms of an expression.” 

Thus: 2 X — 3 becomes 3 — 2 x when its signs are changed. 

2 X — 3 becomes — 3 + 2 x when its terms are rearranged. 

Note. Changing the signs of the terms of an expression changes 
the value of the expression. Rearranging the terms of an expression 
does not change the value of the expression. 






34 


ALGEBRA 


EXERCISE 18 


Simplify the following fractions, expressing the numerical 
coefficient in decimal form when necessary: 


1 . - 


2 . - 


— 2a 
+ 5 a 

4. 

, (+2)(-3) 

- 12 

7. 

(- 7x){-3y) 
— 84 xy 

-36 

5. 

(-3)(+4) 

8. 

5 6(— 6 c) 

— 66 

- 24 

(- 2c)(- 156) 

+ c 
— 4c 

6. 

(- 2)(- 12) 
(- 15)(+ 4), 

9. 

{-5x^K-7f) 
(+2y)(- 7x2) 


10. Reduce to lowest terms 


(x - y)iy - z) 
2 yz{z - y) 


Solution. 1. Change the signs oi y — z. This makes it z — y and 
changes the sign of the numerator and therefore of the fraction. 


2 . 


{x - y){y - z) 


(x - y)(j^^ 


X — y 
2yz 

By changing the 


2yz{z - y) 

3. This is an inconvenient form of the result, 
signs oi X — y, we change the sign of the fraction. 

_ X — y _ y — X 
2yz ~ 2yz 

. {x - y){y - z) ^ y - X 
2yz{z - y) 2yz 

Similarly reduce the following fractions to lowest terms: 

2 +t- 


11 . 


12 . 


13. 


14. 


16. 


a — 2b 
a? 

16. 

(a - hf 

(6 — ay 


21. 

y2 _ ^2 

17. 

(m — n){n — 

■ x) 

22. 

bz - by 

(x — m)(n — 

m) 

ax — b a 

18. 

2 r‘‘ - 2 


23. 

15 - 3x 

2 r{s — r)2 


m2 — n? 

19. 

3x - 15 


24. 

(n — m)2 

20 - 4 X 


c2 - d? 

20. 

S st — rt 


25. 

d? - (^ 

r2 — 9 ^2 



<2 + ^-6 
6 m — 10 
10 — — 3 

5 m — 15 
3 + 2 m — 

4 a — 3 ax 
3 x2 -f 2 X - 8 

^ ^3 
































FRACTIONS 


35 


48, 

Example. Simplify 


Changes in signs in multiplication and division. 

— 1 I — X 


Solution. 1. 


^ X — x^ 
— I 1 — X 
X — 3 


=[- 


Check. When x = 2, 

^ 3 
4 


6 + X — 

(x + l)(x-^) 

{$^){x + 2) 

- 1 


X — 3 

{x + l){x-l) X- 
(3 - x){2 + x)' I- 


]■[- 




^ + X — x^ 

- 1 3 


I — X 

x-3 


Also 


■ a ; +^2 

re + 1 _ 

re + 2 " 


3 

re 


2 + 1 
2 + 2 

3 

4 


EXERCISE 19 

Simplify: 


re^ + 2 re + 1 

1 — X 

6 . 

Is 

1 

•1 

+; 

x^ — 1 

1 + X 

2 — a a? — 3 a 2 

m^ — 2 m + 1 

^ 1 — m 

7. 

X — 3 a ^ 3 a — X 

m^ — 1 

' m + 2 

4 a — X ’ x2 — 7 ax + 12 

re^ — 2/2 

y - 2x 

8 . 

13 X - 2 x2 - 15 . X - 5 

— 2 xy + 

y + X 

9-4 x* ■2x-3 

a2 - 4 0 + 4 

1 

9 . 

(1 - a)* 2 - a 

6 a — 15 

10 - 4 a 

1 

1 

4 — X 

- 3 - 

10. 

r — 2 . 4 — 

6 + X — x2 X 

- 2 

6r-3 ■ 2 - 3r - 2r^ 


c2-2c+l.c2-8c+7 
1 - c2 ‘ 7 + 6 c - c2 

12 - y)Qj - ^ (y - ~ 

y — X ' z — X 
a — b (c — b)(c — a) 

{b — c){a — c) b — a 

6 — S m 6 — 2m , 2 w? ~ S 
5 - lOm ' 6m + 3 ^ 4 m* - 1 
a - 2a? l+4a + 4a* . 2 + 2o-4a2 
4 (t® - 1 ' 3 a ■ 6 a - 6 o* 










































36 


ALGEBRA 


Remedial Instruction and Practice, Pages 36 to 38 

49. Addition and Subtraction of Fractions. 

To add fractions which do not have a common denominator 

we first change them to equal fractions with a common denomi¬ 
nator. 

Thus: 


2 + l_l 

3^2 4 


12 ^ 12 


3 11 

12 ’ 12 


12 is the lowest common denominator (L. C. D.) of the 
fractions. 

2 8 

The change from - to — is based on the following principle, 
o lZ 

50. The second fundamental principle of fractions. Both 
terms of a fraction can he multiplied by the same number without 
changing the ralue of the fraction. ■ 

51. Combining two or more fractions. 

X y z 
6 8 ~ 3* 


Example 1. Find pr + h ~ u 


Solution. 1 . 

6 = 2-3: 

: 8 = 23; 3 = 

3 

.•. the L. 

C. D. of 6 

, 8, and 3 = 2^ 

• 3, or 24. 

[The manner of forming the L. 

C. D. is described in Rule 1, b, page 37.] 

9 9J. -i- A — A 

X 

4: • X 4:X' 


^ O - TC 

’■ 6 ' 

■ 4^’ ^ 


cc 

II 

00 

. y 

“ 8 ■ 

3 • ?/ ^ y 

- - or —- 

3 • 8 24 

This step should be 
done mentally. 

9A • Q — Q 

z 

8 • z 8 z 


O — o 

3 ' 

" 8 - 3’ '''' 24 ^ 



3. 


y 


4 rc , 3 ^ 


6^ 8 3 24 24 24 


8 z 4 a: 4- 3 ?/ 
or - 


8z 


24 


Example 2. Find H- 


b c 

x^ xy if 

Solution. 1. The L. C. D. is x^y^. [See Rule 1, h, page 37.3 
2. xhf ^ x^ = or etc. 


y. 


Q ^ I _ E ^ eL _l _ -- 

x^ xy y^ x^y^ x^y^ x^y^ 


a mp 

-, QT« - 


cx^ aif + bxy — C3p 

-^--4 — 






FRACTIONS 


37 


52. Rule for adding and subtracting fractions. 

1. Change the fractions, if necessary, to equal fractions which 
hare their lowest commo7i denominator. To do this: 

(a) Factor completely each deniominator. 

(b) For the L. C. D., form the product of each of the prime 
factors which appears in the denominators, using with each the 
largest exponent it has m any denominator. 

(c) For each fraction: divide the L. C. D. by the denominator 
of the fraction; multiply both terms of the fraction by the quotient. 

2. For the numerator of the result, combine the numerators ob¬ 
tained in part c of Step 1 in parentheses, preceding each by the 
sign of its fraction. 

3. For the denominator of the result, use the L. C. D. 

4. Simplify the numerator; reduce the fraction to lowest terms. 

Example. Simplify | 


Solution. 


2 . 

3. 


— 6 a; + 8 — 16 

^ 6 _ 5 

{x — 4)(x — 2) {x — 4)(x + 4) 
The L. C. D. = (a; - 4)(a: - 2)(a; + 4). 

L. C. D. ^ (x - 4)(ic - 2) = rc + 4 

6 6{x + 4) 

(x - 4)(x - 2) “ L. C. D. 

L. C. D. ^ (x - 4)(x + 4) = X - 2 
5 _ 5(x - 2) 

(x-4)(x-|-4) L. C. D. 


Hereafter do 
this step 
mentally 
as part of 
Step 4. 


‘ ’ x^ — 6 X + 8 x^ — 16 

6(x + 4) 5(x - 2) 

(x - 4)(x - 2)(x + 4) (x - 4)(x - 2)(x + 4) 
_ (6 X + 24) - (5 X - 10) 

(x — 4)(x — 2)(x + 4) 

_ 6x + 24-5x+10 x + 34 

“ (x - 4)(x - 2)(x + 4)’ (x - 4)(x - 2)(x + 4) 


6. 



















38 


ALGEBRA 


EXERCISE 20. REMEDIAX PRACTICE 
Perform the indicated operations: 


6 . 


6 a 

a 

6. 

3^ 

5< , < 

11. 

2a:+ 1 

3 a: — 1 




—r+x 

4 X 

5 ^ 

3 

7 

14 ^ 2 

3 X 

5r 

3r 

7. 

3 w 

w 2 w 

12. 

X — 1 

x^ — 1 

4 

8 

2 X 

1 

1 

?! 

2 X 

1 

00 


3 X 

8. 

X — 1 

x+l 

13. 

X + Sy 

2x - y 

6 a 

4 a 

2 

4 

x^y 

xy^ 

4 y 

5 y 

9. 

^ + 2 

1 

I 

14. 

2b - c 

c -j- a 

15 6 

■ 66 

3 

6 

be 

ac 

4 c 

5 c 

10. 

2a — 

3 3a - 1 

15. 

to 

1 

X - Sy 

9 a 

m 

8 

6 

4?/ 

6 X 


16. 

17. 

18. 

19. 


20. a - 4 


2 + 11 g 
3 a 


21. a — -;—r "h o 


22. 3 a 


5 a + ^ 

3 - 7a 
2 a - 3 


+ 1 


30 . 


2x + 2y 

+ 

^ 3a^ - 

3t/ 

23. 

y^ — xy * xy — x^ 

36 

56 


24. 

a a 

1 

1 

CO 

5a+ 6 


— 4 m — 16 

3 

4 


26. 

2 — .T 2 + a: 

4 p — 6 

15 p — 

12 

2 + a: 2 - a^ 

3 a 

5 


26. 

4p^+l 2p— 1 

a2 - 9 

a — S 


4 _ 1 2 p + 1 


27. 


2 6 -}- a 


a - b - b‘^ 


28 . - 4 -^+ ^ 


2 a 


a b a — b 


¥ 


29. 


b 


2a — 26 3a-t-36 


— 6 r + 9 + 4 — 21 

a d" 1 


a — 4 ^ a 3 


a^ — a — 6 a^ — 4a4-3 a^+a — 2 


31 . 

















































FRACTIONS 


39 


New Topics, Pages 39 to 43 

53. Combining fractions involving changes in signs. 

17 1 c- y , X 2 2 - X 16 

Example. Simplify— +^ + 


Solution. 1. Rearrange all expressions in descending powers of x. 

. X + 2 2 - X 16 

■■x-2“^x + 2'^4-x2 

X + 2 X — 2 16 


2 . 


X - 2 X + 2 
(x + 2)(x + 2) 
x2 — 4 

(x^ + 4 X + 4) 


(By § 46) 


(x - 2)(x - 2) 
x2 — 4 
4 X + 4) - 


16 


(x2 


x^ — 4 
16 


x^ + 4 X + 4 


x2 - 4 
x^ + 4 X 


16 


8x - 16 
x^ — 4 
8(x - 2) 


x^ — 4 


8 


X + 2 


Check: When x = 1, ^ ^ ^ 


+ 


16 


X — 2 X + 2 

A. + 1+16 
-1^3^ 3 
- 3 + i + 5i, or 2i. 


Also 


X + 2 3 

= 2i 


EXERCISE 21 


1 , X 

1 .- 1 - 2 - 2 

X — y — x^ 

2 1 +-j- 

a a 

3. *7 

— rs s" — rs 

4 _1_+ _i_ 

2mt - 4t &t - Zmt 

2 'iy + X 

X - y'^ - 


6 


6. 

1 ' 1 - 

a 

— 1 

7. 

a h 



CL h CL 

b 

1 

8. 

r r 


r 

r + 2 2- 

r 

1 

9. 

2 


3 

r2 — r — 6 

9 

— 

10. 

1 


2 

— 4 X + 4 


6 — 3 X 














































40 


ALGEBRA 


54. A complex fraction is a fraction having one or more 
fractions in either its numerator or denominator, or both. It 
is merely another way of indicating division. 


Example. Simplify 


62 

a - 

a 

— — 


Solution. 1. 




2 . 


or 


a2 + 62 


(P -j- 62 

Note. A complex fraction can be simplified by multiplying iU 
numerator and denominator by the L. C. M. of the denominators of 
the fractions in its numerator and denominator. In this example, 
multiply by 


EXERCISE 22, a 


Simplify: 
3 


1 . 


2 . 


4 - 
2 - 


-I 

3 + ? 

o 


2 - 


10 


4 . 


5 . 


I ^ 
"+6 


m + 




7 . 


8 . 


3a - - 

(Continued on page 41.) 


2 - 

1 

X 

1 + 

1 

3 a 

a — 

1 

9 c. 



4 - 

1 

t 















FRACTIONS 


41 


10 . 


11 . 


12 . 


13. 


r 


h 

^"2 

2 

" - 4 

X _'l 

3_4 

3 6 

c . d 

d _ c 

c d 

d G 


Simplify: 

‘■(t- 


14. 


15. 


16. 


17. 


a + h 


X + 


2?/ 


a + 6 


X — 5 4 — 

X 


X + 1- 

X 




y - 1 


1 - 


2s 


r 3 ^ 
S s r 


18. 


19. 


20 . 


21 . 


4^ 

9 

_ 1 

X 


2 X - 1 


2 a 
a — b 


- 1 


a — h 


4 - 


y + 1 


16 + 


EXERCISE 22, b {Optional) 


^ /-I 4- 

h d) V9 2/ x/ ■ V6 2 / 2 x) 


3 + 5£z4 

^ ..2 _ Q 


2 . 


X" 


X + 


22 / 


3 + 


X — 3 
2d? 

c d 


1 + 
1 + 


X - 2y 
2x + 2 
x^ — 1 


^ cd — d? 


a + 


c d 
62 


2 - 


- 4 


2x - 2 

y y 


a + 


2 a6 + 6^ 


8 . 


y + 2 y 


y 


+ 


y 


y + 2 2 / - 3 


a — h 










































42 


ALGEBRA 


55. A mixed expression consists of one or more integral 
terms plus or minus one or more fractions. To simplify such 
expressions, follow the rule on Page 14. 

Example. Simplify -i- — 2 + 


Solution. 

2 . 

3. 


+ 1 ) ^ ^ _ 

j3t; G^-<T)(a - 1) 

o -f" 1 


a — 1 

EXERCISE 23 

Simplify by performing the indicated operations: 




1 

2 . 

3. (2 - 




^ _y 

9 y X 

a: + 6 




\ ^ / 6 xY \ 

7 \i^'+3y/ 




3(2 T 


(? t 4 ) *{' - S ) 

“■ (' + i) • (m) * + j) 

... + + 


!) * (, + 

«/ \ + a / 


12 . 1 + 


%){ 


X^ 


1 + 


2xy 


x^ — xy y‘ 


^ + yY 

x‘^ - y^ 


Note. Additional Examples appear on p. 239. 






















FRACTIONS 


43 


56. Dependence of a fraction on its terms. 

When the terms of a fraction are changed, then, of course, 
the value of the fraction changes. 

Example. Consider the fraction - and let x have in order 

the values 2, 3, 4, 5, etc. Thus we get the fractions J, J* 

Clearly as x increases, i decreases. 


EXERCISE 24 

1. If the denominator has a fixed value, and the numerator 

increases, then the value of the fraction_ 

2. In the fraction if x is doubled then - is_ 

X X 

y . . y . 

3. In the fraction if y is trebled, then ^ is_ 

4. If X increases, then 1 + i_ 

X 


6. If y decreases, then 1- 

y 


6. If X increases, then 


1 + ic 
1 — a: 


7. The average of two numbers is 


a -]r h 


If 


a increases 


while b remains fixed in value then the average_ 

8. 6 = -^ is a formula from geometry. If A has a fixed 
value, and h increases then b _; if h is doubled then b 


9. C = - 


E 


is a formula met in the study of electricity. 


r + R 

If E and r have fixed values, and R increases, then C 

1 


10. If X increases, then- 

X 














44 


ALGEBRA 


EXERCISE 25. CHAPTER MASTERY-TEST 


1 . Reduce 


X 2 
0 ^ 2-4 


„ 2 X \ 1 „ 

27T2 = 2- 


3. Reduce 


7. Find 




2 xy x^ 


4. Reduce 
6 . Find 
6 . Find 


- a(c — h) 
b{b - c) 
m . m 


5 a: - 15 ■ 7 a: + 21 
6 5 


0 :^ — 4 (a: — 2)^ 


a: + 1 


8 . Find 


a:^ — 3 X + 2 2 

9 x^ — 1 x^ + 4 X 


8 


9. Find 3 m — 4 

c2 - 9 


16 X 3x + 1 

9 — 16 


10 . Find 


3 m + 4 

— 4 + 2 c 


3c — 6 c^ — c — 6 


3c - 3 


^ - S) 

12 . C = —Find C if F = ^; r = = 1; „ = 3 . 

r H- 


13. h = —If V has a fixed value, how does h change 

when b increases? 

2 A 

14. c = —j - b. If A and h have fixed values, how does 

h 

c change when b increases? 


16. Simplify 

17. Simplify 


3 



x^ + 3 X + 9 
x3 - 27 


X^ — IJ^ 

16. Simplify - j 

^ y 

x‘‘ — Z X x^ — Z X 
6 x+ 18 " 3 - 27 

























IV. FIRST DEGREE EQUATIONS IN ONE UNKNOWN 

57. The following test will help you discover what you still 
know about solving simple equations and guide you in the study 
of this chapter, which is almost entirely review material. 


DIAGNOSTIC TEST 8 


1 . The expression 3a^ — 5 = 2 a: + 4 is an_; 

3 a: — 5 is called its__; 2 x + 4 is called 

its_ 

2 . If X — 5 = 6 , then x =_This is secured by 

_to both sides. 

3 . If X + 5 = 9, then x =_This is secured by 

_from both sides. 

4 . If 3 X = 15, then x =_This is secured by- 

both sides by- 

6 . If ^ = 3, then x =_This is secured by- 


both sides by- 

If3x — 5 = 2x + 4, does x = 9? 


6 . 


If X does equal 9, then x is called the 

and is said to__the equation. 

If 2 ?/ + 4 = i/+6, does ?/ = 3? (Yes, or No) 

,• ^ I ^ o t o 


(Yes, or No)- 

_of the equation. 


7. 

8 . 
9. 

10 . 

11 . 

12 . 


lrz?/-i-^ = yq-o, Qoes y = o: v ui - 

Solve the equation 4w+7 = 3m + 2. Check. 

Solve the equation 7<+3 = 5^ + 4. Check. 

Solve the equation llx — 7 = 6x + 2. Check. 

.XX 13 3 X 1 

Solve the equation 3 “ 2 ^ ^-5" 

Solve the equation 5(x — 2) = 3(x— 1) — 7. Check. 
45 

















46 


ALGEBRA 


58. An equation expresses the equality of two numbers. The 
two parts are called the sides or members of the equation. 

59. An equation is satisfied by a set of values of the literal 
numbers in it if the two sides become numerically equal when 
these values are substituted for the literal numbers. 

60. An identity is an equation in which the two sides may be 
made to take exactly the same numerical value or the same 
form by performing the indicated operations. 

(a) 10 + 2 = 15 + (2 X 4) — 11 is a numerical identity. 

(b) S x{a — b) = 3 ax — 3 bx is an identity. It is satisfied by 
any set of values of a, b, and x. 

61. A conditional equation is an equation involving one or 
more literal numbers which is not satisfied by all values of the 
literal numbers. 

Thus, 3x — 4 = x-\-Q is an equality wily when x = 5. 

The word equation usually refers to a conditional equation. 

62. An equation like 3a: — 5 = Jx+7isa first degree or a 
simple equation. Observe that x has exponent 1 and is not in 
the denominator of a fraction, or under a radical sign. 

63. If an equation has only one unknown number, any value 
of the unknown which satisfies the equation is called a root of the 
equation. 

64. The following axioms are used in solving a simple equa¬ 
tion. 

(a) The same number may be added to both members of an 
equation without destroying the equality. 

(b) The same number may be subtracted from both members of 
an equation without destroying the equality. 

(c) Both members of an equation may be multiplied by the 
same number without destroying the equality. 

(d) Both members of an equation may be divided by the same 
number without destroying the equality. 


FIRST DEGREE EQUATIONS 


47 


65. In this text, symbols A, S, M, and D are used to abbre¬ 
viate the explanations of solutions of equations. Thus: 

As: means “add 5 to both members of the previous equation.” 

S-zn- means “subtract — 3 n from both members of the 
previous equation.” 

M_i: means “multiply both members of the previous equa¬ 
tion by — 1 .” 

D 4 : means “divide both members of the previous equation 
by 4.” 

Example 1 . fx — 5 = 3a: — f 


Solution. 1. To eliminate the fractional coefficients: 

2 o ^ ^ 

Me: - 6 • 5 = 6 • 3 X - 


2. 4 X - 30 = 18 X - 9 

3. A 30 : 4 X = 18 X + 21 

4. S 18 .: - 14x = 21 

rr . _ 21 3 

5. D_i4. X 14* 2 


Check. 

Does - 1 - 5 = - ? - I? Yes 


EXERCISE 26. REMEDIAL PRACTICE 
Solve and check the following equations: 


1. 5x-2 = 2x+l 

2. 8x-9 = 2x — 6 

3. 10r-3 = 5 + 6r 

4. 4y + 3 = 19-7/ 

5. 5y-l = 4y + 5 

6. 8^ — 3 = 25 + 4^ 

7. 11 6 - 6 = 1 - 10 6 

8. 21w — 6=1 — 7^ 

9 . At - 3 = S - t 

10 . 15w-5=7-9w 


11. 10 p — 1 = — 5 p + 4 

12. 7 X — 1 = 3 X + .6 

13. X - f X = f 

14 . iy + iy = ^ 

16. it = it + i 

16. ix + ix = i + ix 

17. c - = i c 

18. im — = — 

19. i{x — 1) — 7 = 9 

20. 5(< + 6) - 10 = 50 



48 


ALGEBRA 


66. There are three mechanical methods of solving a simple 
equation the use of which is optional. 

(а) Transposition. A term may be transposed from one 
side of an equation to the other, provided its sign is changed. 

Proof. 1. Let x a = b. 

2. Sa. X = b — a. Axiom (b), § 64 

The result is the same as if + a, with its sign changed, were 
taken from the left side and placed on the right side of the 
equation. 

(б) Cancellation. A term which appears in both sides of 
an equation may be cancelled. 

Proof. 1. Let x a = b a. 

2. So. X = b. Axiom (6), § 64 

The result is the same as if -]r a were simply dropped from 
both sides of the equation. 

(c) Changing signs in an equation. The signs of all of the 
terms of an equation may be changed. 

Proof. 1. Let ax — b = c — dx. 

2. M_i. — ax b'= — c + dx. Axiom (c), § 64 

The result is the same as if the signs of all the terms of the 
equation were simply changed. 

Remember, these rules need not be used. They are used 
commonly by mature mathematicians, so your teacher may 
wish you to become acquainted with them. Use them or not 
as your teacher directs. 

Example. Solve the equation 

5-6x-lla:=13-na:-4x 

Solution. 1. 5 — Q X — 11 X 13 — 11 X — 4: X 

2. Cancelling 11 x: 5 — 6x = 13 — 4a; 

3. Transposing 5 and — 4a;: — 6a; + 4a; = 13 — 5 

4. Collect terms (C. T.) — 2 a; = 8 

5. Changing signs: 2 a; = — 8 

6. a; = — 4 


FIRST DEGREE EQUATIONS 49 

67. Solving equations with fractional numerical coefficients. 

Example. Solve the equation 

5 ?/2 - 4 ^ 17 _ 10 2/2 + 9 1 / 

5 2 ” 10 

Solution. 1. First we must eliminate the denominators. We do 
this by multiplying by the L. C. D., namely 10. 

3. - 8 2/ - 85 = Vd-tp + 9 2/ 

4. Assj Sgj/ 17 2/ = 85 

5. D_i7 y = — ^ 

Check Does 5 ' 25 + 20 _ H ^ 10 ^ 25 - _45 

Does ^ ? Does 29 - 8§ - 20i ? Yes. 

0.^10 


EXERCISE 27. REMEDIAL PRACTICE 

Determine by substitution which of the numbers 1, 2, — 3, 
and \ are roots of the following equations; 

1. 5a:-2 = 2z+l 3. 8a:-9 = 2a:-6 

2. + 2 .T = 4 4. 2 3 .t - 2 = 0 


Solve the following equations: 
6. 3(n - 2) = 14 - 

6. \ X = ^ X — I 

7. = 

s. iy - I = iy - 

9. -f-z — |- 2 ; = |- — ^2 

15. + 5) - 4 : 


10. 4m — |-J = |-^+Jm 

11. i{x - 2) = i(7 - .t) 

12. iix + 7) = K4 - x) 

13. 4(32/-10) =1(10-32/) 

14. iV(l - ^) = if - 4 
i{x - 10) 


16. -poim - 1) - 4(5 m + 7) = -if 

17. 4(5 - 3) = 4 5 - 4(3 ^ - 7) 

18 . 1 ( 2 / - 1 ) = 4 ( 2 / “ 3 ) - 








50 


t 


ALGEBRA 


68. Solving fractional equations. 

Example. Solve the equation 

3 1 +4:X - 4 

2 X + 1 “ 4 - 1 

Solution. 1. Eliminate the denominators by multiplying by the 
L. C. D.; that is M( 4 a:»_i) 


(2x-l) / 3 \ , .X ,__l+4x-4x! 

2. /. 6 X - 3 - 4 x2 + 1 = 1 + 4 X - 4 x2 

3. A4a:2 6x — 2 = 4x+l 


4. 


2 X = 3, or X = - 


Check: Does | “ 1 = ^ 9 • !. i " " ? Does - ^ ? Yes. 


In the check, observe that 2x + l “^( 2 /’^^ = 3 + 1, or 4, 


Example 2. Solve the equation 

2t+ 7 3^-5 17^+7 

6^-4 9 ^+ 6 ~ 9 / 2-4 

Solution. 1. First factor the denominators: 

2^ + 7 3f-5 _ 17i + 7 

2(3 i - 2) 3(3 i + 2) (3 ^ - 2)(3 « + 2) 

2. Multiply both sides by 6(3 ^ — 2)(3 i + 2). To do this, imagine 
the multiplier as standing in front of each fraction in turn. 

For example, ^(3 ( + 2), • = 3(3 1 + 2)(2 t + 7). 

In this way you will get: 

3(3 i + 2)(2 ^ + 7) - 2(3 i - 2)(3 f - 5) = 6(17 t + 7) 

3. .*. 3(6 t^ + 25t+ 14) - 2(9 t^ - 21 t + 10) = 102 t + 42. 

4. .*. 18 ^2 + 75 « + 42 - 18 t^ + 42 ^ - 20 = 102 f + 42 

5. 117 ^ + 22 = 102 ^ + 42 

6. 15 i = 20, or f ^ 

4 


Note. If the answer found makes any denominator zero, then it 
cannot be a root of the given equation for division by zero is not per¬ 
mitted. (See § 20.) 














FIRST DEGREE EQUATIONS 


51 


EXERCISE 28. REMEDIAL PRACTICE 
Solve and check the following equations: 


35 


2 s 3 2 s 

5c — 4_ lOc+9 

5 “ To 

2 - X ^ _4_1 

15 x 6 
X — 4 


iZ 

2c 


5 X 

X — 5 _ 

X + 3 X + 6 
4 3x + 1 


X + 1 
2 


x^ — 1 
7 + m 
m — 3 


x2 - 1 

9 - y 

y'' - y 

1 

X — 1 
m + 4 
m — 5 
3 X 


9. 

10 . 

11 . 

12 . 

13. 

14. 

16. 

16. 


3 a 


8 


a — 3 5 a 

9x4-2 


3x - 1 
4 _ 

X — 4 
4 

X — 


= 5 


15 ^ 
2x - 3 
X — 1 


X — 3 
5 


5 4 + X 


3 X — 9 
= 0 

2 


4 X ~ 2 X — 2 x(x — 1) 


X 4- 3 

2 - X 

z 

s - 4 

X 


12 


2 4- X 


17. 


18. 


19. 


20 . 


21 . 


22 . 


36 


3z - 12 

3 _ X 4“ 8 

X - 6 “ 2x - 12 X -h 2 
4 X 2 4- 3 X 


X 4- 3 
3 


X — 3 
2 


X — 3 x^ — 5x4-6 X — 2 
Note. See Note, p. 50. 

X _ 3 X 4- 10 

x4-2 x2 4-5x4-6 
5 ^ n 

4x - 2 6 

X X 4- 25 


X 4- 3 
3 


2 X 
X 


4- 


X 4- 7 X 4- 1 x^ 4- 8 X 4" 7 
6 x2 4- 14 X - 4 X 


x4- 2 


4- 5 




















































52 


ALGEBRA 


EXERCISE 29 {Optional) 


1 . 

2 . 

3. 

4. 

6 . 


Harder Fractional Equations 
Solve the following equations: 
a: + 5 4 _ 

=^"^ + ^ + 3 = 


6 . \2 = 5 -^ + 1 

1 — a: 3 — a; 

2 a: 4 a: + 5 3 


2a:-3''^ 2a:- 3 

15 a:2 - 5 a: - 8 


3 a:^ + 6 a: + 4 
3 4 


= 5 


r — 2 2 r — 1 r+4 

9 _^ ^ 1 

3c — 5 c — 2 c — 3 


10 . 


3a: — 4 6a:— 1 3a: — 4 

6a:+5 2 _ 3a: 

2 a:^ — 2 a: 1 — a:^ a:^ — 1 

3 m 7 _ 

2m — 5 3m+l 

4 a: 4 1 

2.r-6 5 a:- 15 2 


11 4 . = 2 - 15 

’ 22 / 4-3 3 - 21 / 4 2 / 2-9 

12 ^ ^ ~ 14 ^ 2/ + 3 _ 2/ - 2 

*2/^ 4-32/ - 28 2/+7 2/-4 

a4-2 2a — 3 26 — 

a - 4 ~ a 4- 3 "" a^ - a - 12 

14 2 z 4~ 1 _ 2 s 1 ^ 18 z 4" 34 

2-8 s -h 6 ~ s2 - 2 s - 48 


16. 



a: 4- 2 _ 6 a: - 5 
2a: -f- 5 ~ 6 


16. 

17. 

18. 

19. 


y _ 9 // - 2 2 2/2 

3 9 3 2/ - 4 

2 a: -f 7 _ a: 4- 6 ^ 5 a: - 4 
14 7 3 a: 4- 1 

5/4-1 4/ 3/-2 7 

8 5< + 8 - 3 5<+ 8 

3x-2_ 36- 4x 2 + 3.T 

x +3 - 9 


3 — a: 


CO icq 




















































FIRST DEGREE EQUATIONS 


53 


69. Solving literal equations. An equation like 
a{bx — a) = b{a — bx) 

is a literal equation. In such equations, we assume x is the 
“unknown.” 

Solution. 1. a{hx — a) = h{a — hx) 

2. .•. abx — = ah — ¥x 

3. Get the terms containing x on one side, and the remaining terms 
on the other side. 

.•. abx + ¥x = a^ ab 

4. Factor out the coefficient of x. 

(ab + b^)x = + a6 

a^ + ab 


5. D(o6 + 6^) 


6. Reduce to lower terms: x 


ao + 
a{a + b) 


b{a + 6)’ 

Does a{a — a) = b{a — a)? Does a • 0 = 6 • 0? Yes. 


Check. Does a(Jb • r — a) 


, or X 
? 


EXERCISE 30 

Solve the equations: 

1. 5{x - Sb) = 3x - lib 

2. ax — ac — bx — be 

3. r{x — r) = s{s 2 r — x) 

4. c{x - c2) = d{(P - x) 


6 . a{3 bx - 2 a) = b{2a - 3 bx) 

7. (x — a)‘^ 2 x^ = 3 x(x — a) 

8 . rri^{x - 1 ) — rnix — 2 ) = 1 

9. 3 0 !^ — 3ax = 10b‘^ - ab-5bx 


6. ax — 

= - b{b 

+ x) 10 

. — nx = mx — 


ax — b 

a — 

b bx a 

11. 

bx 

abx 

ax 


X n 

X — n _ 

2(m + nY 

12. 

X — m 

X m 

x^ — 


X — b 

X + b 

4 a^ 

13. 

X — 2 a 

x+ 2a 

11 

1 

14. 

2x 

3 a 

2 x^ -\- ax 

1 

CO 

x+2a 

3 x^ 5 ax 2 
















54 


ALGEBRA 


70. Algebraic translation. Every algebraic expression con¬ 
taining one or more literal numbers is a formula for a number. 

Thus, is the formula for the difference of the squares of two 

numbers, although it may appear incomplete. By letting d = — y^, 

a formula in customary form is secured. This however is unnecessary. 

EXERCISE 31. REMEDIAL PRACTICE 

Express in symbols; or, write the formula for: 

1. The sum of the cubes of two numbers. 

2. The larger part of w if x is the smaller part. 

3. The number which is 5 more than x. 

4. The number which exceeds y by 3. 

6. The result of diminishing t by 8. 

6. The excess of 15 over x. 

7. The amount by which 18 exceeds twice a given number. 

8. One half of the sum of two numbers. 

9. (a) The distance a man travels in t hours at rate 20 m. p. h. 
(miles per hour). 

(6) The distance a man travels in t hours at r m. p. h. 

10. The time required to go D miles at the rate r m. p. h. 

11. The rate travelled by a party which went D miles in t 
hours. 

12. R per cent of P dollars. 

13. The complement of x degrees; the supplement. 

14. The number of cents in x nickels and y dimes. 

16. The difference between one eighth of a certain number and 
one twelfth of the same number. 

16. The integer (whole number) next larger than the integer 
n (or consecutive to the integer n). 

17. The even integer consecutive to (and larger than) the 
even integer n. 


FIRST DEGREE EQUATIONS 


55 


71. Some problems can be solved by expressing the condi¬ 
tions in the form of an equation. The ones on the following 
four pages are commonly required as part of the third semester 
algebra. Obviously, you should read the problem carefully, 
make certain that you understand all words in it (like exceeds, 
consecutive, integer), understand what you have given and what 
you are to find. 

You must learn how to translate from English to algebra. 
Some of this you have done on page 54. Some problems you 
can translate quite directly. 

Example 1. Twice a certain number exceeds 18 by 6 
becomes 2n — 18 = 6 

Example 2. Separate 85 into two parts such that the quotient 
is 2 and the remainder is 5, when the larger is divided by the 
smaller. 

Let the parts be I for the larger and 85 — I for the smaller. 

—^ = 2 + —^ 

85 - ^ ^ 85 - Z 

EXERCISE 32. REMEDIAL PRACTICE 

1 . The denominator of a certain fraction exceeds its numera¬ 
tor by 5. If the numerator be decreased by 3, and the denom¬ 
inator be increased by 1, the resulting fraction has the value f. 
Find the fraction. 

2. Separate 135 into two parts such that, when the larger 
is divided by the smaller, the quotient is 3 and the remainder 
is 23. 

3. Find three consecutive integers whose sum is 141. 

4. Find three consecutive even integers whose sum is 234. 

6. Find three consecutive odd integers such that twice the 
square of the largest exceeds the sum of the squares of the other 
two by 208. 





56 


ALGEBRA 


72. Uniform motion problems are based on the assumption 
that some object is moving at the same rate for a specified time. 

In such cases, the distance = rate X time, or d = rt. From 
this equation, r = d t and t = d r. 

Example. An automobile party has 5 hours to spend on a 
trip. They decide to go as far as they can, traveling at the 
average rate of 20 miles per hour going, and at 30 miles per 
hour returning. How far can they go? 

Solution. 1. Let d = the no. of miles they can go. 



Distance is 

Rate is 

Time is 

Going 

Returning 

d mi. 
d mi. 

20 mi. p. h. 
30 mi. p. h. 

d -7- 20 hr. 
d u- 30 hr. 



Solving this equation, d = 60 miles. 

Check. It takes 3 hr. to go, and 2 to return. 

EXERCISE 33. REMEDIAL PRACTICE 

1. If two automobiles start toward each other, at the same 
time, from points 175 miles apart, one going 22 miles per hour 
and the other 28 miles per hour, how soon will they meet? 

2. In Example 1, if the first car starts at 8 a.m. and the 
second at 9 : 30 a.m., at what time will they meet? 

3. A man made a trip of 190 miles in 7 hours, part at the 
rate of 15 miles per hour and the rest at an average rate of 
30 miles per hour. How far did he travel at each rate? 

4. The rate of one train is 8 miles per hour less than that of 
a second train. If the former requires five hours to go the same 
distance that the latter goes in three and two-thirds hours, 
what is the rate of each? 

Note. Additional problems appear on p. 240. 









FIRST DEGREE EQUATIONS 


57 


73. Problems about interest on money. {Optional) 
EXERCISE 34 

1 . A man has $3000 invested at 8% and $4000 at 7%. How 
much must he invest at 5% to make his total annual income be 
6% of his total investment? 

Solution. 1. Let x = the no. of dollars he must invest at 5%. 


Am’t Invested 

At Rate 

Interest 

3000 

8% 

240 

4000 

7% 

280 

X 

5% 

.05 X 

(7000 + x) 

6% 

.06(7000 + x) 


3. .'. .06(7000 + x) = 240 + 280 + .05 x. 

(Complete and check the solution) 

2. From three investments, a man receives $224 interest. 
The amount invested at 6% is $1000 more, and the amount 
invested at 5% is $400 more than the amount invested at 7%. 
How much has he invested at each rate? 

3. A man has $1800 invested at 5% and $2500 invested at 
7%. How much must he invest at 8% to make his total income 
be 6j% of his total investment? 

4. A man has $8000 to invest. He wants to have his total 
income be 6% on the amount invested. He can invest part 
at 5% and part at 6^%. What must he invest at each rate? 

5. If N dollars are to be invested, part at 8% and part at 
5%, how much must be invested at each rate so that the total 
income will be 6% of the N dollars. 

6 . A man has A dollars invested at r per cent, and B dollars 
invested at s per cent. How many dollars must he invest at 
t per cent to make his total income be n per cent of his total 
investment? 













58 


ALGEBRA 


74. Mixture problems. {Optional, but required by C, E. E. B.) 

EXERCISE 35 

1 . How many pounds of 25^ coffee must be mixed with 40 lb. 
of 60^ coffee to make a mixture which can be sold for 40^? 

Solution. 1. Let n — the no. of lb. of 25^ coffee. 


Kind of Coffee 

The No. of Lb. 

Value 

25^ kind 

n 

25 n 

60kind 

40 

60 X 40^ 

40^ kind 

{n + 40) 

40(n + 40)^ 


3. .*. 25 n + 60 X 40 = 40(n + 40) 

(Complete and check this solution) 

2. How many pounds each of 35^ coffee and 50jzf coffee must 
a grocer mix to make 100 pounds to sell at 40jZ^ per pound? 

3. How much alcohol must be added to 4 gallons of a 30% 
mixture of alcohol and water to make a 50% mixture. 

Note. This means that 30% of the 4 gallons is alcohol and 70% is 
water. 50% of the resulting mixture is to be alcohol. 

Solution. 1. Let n = the no. of gal. of alcohol to be added. 


In the 

The Quantity is 

The Alcohol is 

old solution 

4 gal. 

.30 X 4 gal. 

new solution 

4 + n 

.50 X (4 + n) gal. 


3. .50(4 + n) = n + .30 X 4 

(since the alcohol in the new solution consists of that in the old plus 
the n gallons added. Complete this solution and check it.) 

4. A radiator contains 7j gal. of a 25% mixture of alcohol 
and water. How much of it must be drained off and replaced 
by alcohol so that the resulting mixture will be a 50% mixture? 

Note. Additional mixture problems appear on p. 241. 


















V. FUNCTIONAL RELATIONSHIP 


75. The meaning of functional relationship is easily intro¬ 
duced by some examples. 

Examples, (a) The total cost of the outfits for a football 
team depends upon the cost of the outfit for each of the eleven 
men. 

(b) The wages earned by a workman who is being paid 75jzf 
per hour depends upon the number of hours he works. 

(c) The number of hours required by an automobile for a 
trip of 150 miles depends upon the average rate per hour. 

(d) The area of a rectangle depends upon the lengths of its 
base and its altitude. 

Always there is one number or quantity so related to one 
or more others that any change in the latter causes a change 
in the former. The numbers or quantities involved in such a 
situation are said to be functionally related to each other. The 
essential fact is that a definite value of one of the numbers de¬ 
pends upon or corresponds to each value of the other number, or 
numbers. 

Many important functional relationships cannot be expressed 
by algebraic symbols. Thus, ‘‘the price of wheat depends 
upon the size of the crop.” While true, there is not a mathemati¬ 
cal formula connecting “price” and “size.” 

We are concerned especially with those relations which are 
expressible by algebraic symbols. In this chapter, you will 
learn some of the means of expressing such relations, and will 
learn a little about drawing conclusions from such expressions 
of the relations. 


59 


60 


ALGEBRA 


76. One functional relation expressed in four ways. 

(а) Expressed in words. The wages earned by a man working 
for 75^ per hour depends upon the number of hours he works. 

(б) Expressed by a formula. 

If n = the no. of hr. the man worked, 

and W = the no. of dollars he earned, 

then h = .75 n. 

(c) Expressed by a table. 


If the man works 

1 2 

3 

5 10 

15 

20 

30 

40 

hours 

then his wages are 

.75 1.50 

2.25 

3.75 7.50 

10.75 

15.00 

22.50 

30.00 

dollars 


(d) Expressed by a graph. 

On the graph below, the spaces on the horizontal lines repre¬ 
sent 1 hour, and those on the vertical lines, 1 dollar. 


W 



Each of these means of expressing the relation has some 
advantage. For this reason it is desirable to know how to 
express by each of the other three means a relation expressed 
by one of the four means. In this chapter you will learn how to 
do this when the relation is expressed first by means (a) or (6). 




























































































FUNCTIONAL RELATIONSHIP 


61 


77. Some necessary definitions. 

Function. If one number is so related to another number 
(or numbers) that the former has a definite value or values for 
every value of the latter, then the former is a function of the 
latter. 

Thus, the perimeter of a rectangle is a function of its base and alti¬ 
tude, because the perimeter depends upon them. 

78. A formula (in a narrow sense) is an algebraic statement 
of the rule for computing one number which depends upon or 
is a function of one or more other numbers. In a broader sense, 
a formula is an algebraic statement of the functional relation 
between two or more numbers. 

Thus, P = 2(a + 6) expresses the functional relation between P, 
a, and b. 

79. A variable is a literal number which has any one of several 
arithmetical values during a particular discussion. 

A constant is a number which has some fixed value during 
a particular discussion. 

Thus, if we consider all rectangles of altitude a, then in the relation 
P = 2(a + 6), a is a constant, and P, and b are variables. 


The variable (or variables) in a formula which changes first 
is called the independent variable. 

The variable whose change is caused by the change in the 
independent variable (or variables) is called the dependent 
variable. 

Thus in the formula P = 2(o + 6), if a is constant, and we give 
values to b, and determine the corresponding values of P, then b is 
the independent variable and P is the dependent variable. In this 
case P is a function of b. 

If however we give values to P and determine the corresponding 
values of b, then P is the independent variable and b is the dependent 
variable. In this case 5 is a function of P, 


62 


ALGEBRA 


80. Write the formulas for the following functional relations. 

1. The area {A) of a triangle equals one half the product 
of its altitude {h) and its base (6). 

2. The area of a circle equals tt times the square of its 
radius (r). 

3. The area of a regular polygon equals one half the product 
of its perimeter (p) and the radius (R) of its inscribed circle. 

4. The volume (F) of a pyramid equals the product of one 
third its altitude and the area (R) of its base. 

6. The volume of a sphere equals four thirds of the product 
of TT and the cube of its radius. 

6. The surface (*S) of a sphere equals four times the product 
of TT and the square of its radius. 

7. The amount (*S) to which P dollars accumulates at f% 
compounded annually for n years equals P multiplied by the 
nth power of the binomial 1 plus r. 

8. The sum of the squares of two sides {a and b) of a tri¬ 
angle equals twice the square of one half the third side increased 
by twice the square of the median drawn to that side. 

9. There is a solid whose volume (F) equals one sixth of its 
altitude {h) multiplied by the sum of its bases {b and c) and 
4 times its mid section (m). 

10. The number of gallons (g) in a rectangular tank having 
dimensions L ft., W ft., and H ft. is ^ of the product of L, W, 
and R. 

11. The area of a trapezoid equals one half its altitude mul¬ 
tiplied by the sum of its bases. 

12. The square upon the hypotenuse (k) of a right triangle 
equals the sum of the squares of the other two sides (a and b). 

13. The force (F) equals the mass (m) times the accelera¬ 
tion (a). 


FUNCTIONAL RELATIONSHIP 


63 


81. Expressing functional relationships by a table. 

Example. By the formula I = prt, find the values a, b, c 
and d, in the adjoining table. 

Solution (a) I = 1000 X .05 X 2 = 50 X 2 
= 100 

Solution (b) I = 1500 X .05 X 2 = 75 X 2 
= 150 

Solution (c) 180 = 1500 X .06 t; 90 t = 180; 
t = 2 

Solution (d) 270 = P X .06 X 3; 

.18 P = 270; P = 1500 

The completed table helps you understand that; 

I increases when p increases if r and t are constant; 

I increases when r increases if p and t are constant; 

I increases when t increases if p and r are constant. 

To find the value of any variable in a formula when the 
values of the other variables are known requires ability to solve 
equations. Examples of this kind will appear as the different 
kinds of equations are taught. 

EXERCISE 36 

1 . Complete the following table using 
the formula V = Iwh. 

2 . V depends upon- 

3. If / and w are constant, V increases 

when_ 


4. Complete the adjoining table using 
the formula C = irrh. 

6. If r is constant, C increases when 

6. If r is constant, and h is doubled, 
_ 


V 

1 

w 

h 


20 

8 

5 

1000 

25 

10 


1200 

30 


4 

3600 


15 

6 


C 

TT 

r 

h 


2 2 

7 

14 

5 

440 

2 2 

7 

14 


396 

2 2 

7 


21 


I 

V 

r 

t 

a 

$1000 

.05 

2 

h 

$1500 

.05 

2 

180 

$1500 

.06 

c 

270 

d 

.06 

3 





































64 


ALGEBRA 


82. The vocabulary of graphical representation. 

In the figure below: XX' is called the horizontal axis; YY' 
is called the vertical axis; together they are called the axes; 
the point 0 is called the origin; PR, perpendicular to the 
horizontal axis, is called the ordinate of the point P; PS, 
perpendicular to the vertical axis, is called the abscissa of 
P; PR and PS together are called the coordinates of P. Dis¬ 
tances on OX are considered positive, on OX' negative, on OY 
positive, and on OY' negative. The part of the plane within 
the angle XOY is called the first quadrant; the part within 
the angle YOX' is called the second quadrant; etc. The abscissa 
of P, according to the indicated scale, is 3, and the ordinate is 4. 
The point P is called the point (3, 4). 



EXERCISE 37 

1. What are the coordinates of A^i P? 6r? P? 

2. In what quadrant does a point lie whose coordinates are 
both negative? Whose abscissa is negative and ordinate posi¬ 
tive? Whose abscissa is positive and ordinate negative? 


































































































FUNCTIONAL RELATIONSHIP 


65 


83. Study of first degree functions of x. 

Expressions like x, ^ x, 2 x — 5 are first degree functions of 
X. We shall study first the functions ^ x, ^ x — 5, and | a; + 5. 


For convenience let ?/ = | x. 


Whenx = 

— 10, then y = - 

- 5 

“ X = 

0, 

(( 

y = 

0 

“ X = 

1, 

(6 

y = 

1 

2 

“ X = 

2, 

Ci 

y = 

1 

“ X = 

4, 

iC 

y = 

2 

“ X = 

6, 

ii 

y = 

3 

“ X = 

10, 

(( 

y = 

5 


The graph is the line AB. 



Also \ety = ^x — 5 
When a: = — 10, then y = — 10 

- a:=- 4, ‘‘ y = - 7 

‘‘ X = 0, ‘‘ y = - 5 

X = + 4, 2/ = - 3 

- X = + 6, 2/ = - 2 

X = + 10, 2/ = 0 

The graph is the line EF. 


Also let 2 / = i X + 5 


When 

X = 

- 10, 

then 

y = 

Ci 

X = 

- 4, 

CC 

y = + 

iC 

X = 

0, 

cc 

2 / = + 

6C 

X = 

4, 

(C 

2 / = + 

iC 

X = 

6, 

ii 

2 / = + 

(( 

X = 

+ 10, 

a 

y = 


The graph is the line CD. 


0 

3 

5 

7 

8 
10 


Observe that all three functions increase J unit when x in¬ 
creases 1 unit. This is caused by the coefficient, J, of x. 

AB goes through the origin and the point x = 1, 2/ = i- 

Observe that EF and CD are parallel to AB, and also rise J 
space for each space to the right. 

Observe that EF is 5 units below AB; this is caused by the 
— 5 of the function J x — 5. 

The 2 /-intercept of ^ x - 5 is - 5; it is got by letting x = 0. 

The x-intercept is + 10; it is got by letting y = 0. 

Observe that CD is 5 units above AB; this is caused by the 
+ 5 of the function J x + 5. The y-interceyt is + 5; the x-inter- 
cept is — 10. 






















































66 


ALGEBRA 


84. Summary of facts about first degree functions of the 
form ax and ax + 6 in which a and b are constants. 

1. These functions increase a units when x increases 1 unit. 

2. The graph of every first degree fu7iction is a straight line. 

3. The graph of ax goes through the origin-, and since the 
function increases a units when x increases 1 unit, it also goes 
through the point (1, a). 

4. The graph of ax b is parallel to ax, and is b units above 
or below it according as b is positive or riegative. 

Example 1. What can be said about the functions y = — 3x, 
y = — 3 a: + 5, and y = — S x — 7? Draw their graphs. 


Solution. 1. These functions all in¬ 
crease (—3) units when x increases 1 
unit; that is they decrease 3 units when 
X increases 1 unit. 

2. ^ — 3 a: is the straight line through 

the origin and the point 1, — 3. 

3. ^=—3a:-l-5 is parallel to ^ = 

— 3 a:, and is 5 units above it. 

4. y = — Z X — 1 is parallel to y = 

— 3 X, and 7 units below it. 


Y 



Example 2. Draw the graph of the function y = % x 2 
and discuss it. 


Solution. 1. First draw with a dotted 
line the graph of y = § x. This is a 
straight line through the origin and the 
point X = 1, ^ = |. 

2. ^ = I a; -1- 2 is parallel to y = i x 
and 2 units above it. Draw this line as a 
full line. 

Observe: both lines rise | of 1 unit when 
X increases 1 unit. 



For 2 / = f a: + 2, the y-intercept is -h 2, for ^ = 2 when a: = 0. 
For ?/ = I x 4- 2, the x-intercept is — 3, for x = — 3 when y = 0. 



















































































































FUNCTIONAL RELATIONSHIP 


67 


EXERCISE 38 

Make such statements as you can about the following func¬ 
tions; draw the graph and give the x and y intercepts of each. 


1. 

3x 

6. 4x+ 6 

9. 

i X — 5 

2. 

— 5 X 

6. 3 a: - 5 

10. 

— ^ X -j- 6 

3. 

ix 

7. - 2 X + 3 

11. 

— 1 X — 2 

4. 

- 1 X 

8. - 3 X - 2 

12. 

ixA7 


13. In the formula A hh, \et h = b) then A = 5 6. Now 

A is a_degree function of h. The graph will be a _ 

_ through _ A will increase _ 

units when b _ I unit. 

14. In the formula I = Prt, let r = .05 and t = 2. Then 

I = _Now 7 is a _ of _ 

The graph is a__ through _ __ 

I _units when P increases I unit. 

16. In = C + G, let G = 2. Then *S is a _ 

function of_When C increases I unit, then S _ 

_The graph is a _ It crosses the 

C-axis at C =_It crosses the 5^-axis at = _ 

16. In = rt, let t = 5, then d =_Now 7 is a 

_of_The graph is a _ 

_ through the_For an increase of one unit in 

r, there is an _ _of _ units in_ 

17. F = ma is a formula in physics. When a = 200, then 

F = _Then F is a_function of_ 

18. In p = 2 a + 2 6, let 6 = 10. Then p is a_ 

_function of_ When a increases 1 unit, then p 

_unit. 

19. In / = 2 a + (n — 1)7, let = 10 and 7=2. Then 

I is a _ function of_When a decreases 1 

unit, then I ____ units. 



































68 


ALGEBRA 


85. A formula usually indicates definitely how one number 
can be computed when certain other numbers are given. 

86. Some modern views about accuracy in computation. 

(а) Degree of accuracy in a measure. When we say that the 
radius of a circle is 5, we really mean that it is as nearly 5 as 
we can determine it. The accuracy of a measure is indicated 
by the number of significant figures in it; thus 5.1 implies 
a more accurate measure than 5 or 6, and a less accurate measure 
than 5.16 or 4.75. 

(б) Rounding off numbers. The value of tt is 3.14159. 

3.14159 correct to five figures is 3.1416. 

3.1416 correct to four figures is 3.142. 

3.142 correct to three figures is 3.14. 

When dropping off a final figure 5, computers increase by 1 
the preceding figure if it is odd, but do not increase it if it is 
even. 

Thus, 3.135 becomes 3.14, also 3.145 becomes 3.14. 

(c) Computing with approximate values. The circumference 
of a circle with radius 5 inches is 2 X 5 X 3.14159, or 31.4159 
inches. 

A reasonable result would be “about 31 inches.” Some 
computers, however, would reason as follows: since there is 
only one significant figure in 5, then round off the answer to only 
one significant figure. This gives “about 30 inches.” 

In the following exercise: 

(a) Carry out the computation with the numbers given. 

(b) Give what you consider a “reasonable result.” 

(c) If your teacher wishes you to secure practice in use of these 
new rules of computation, round off your final result so that 
the number of significant figures is the same as appears in that 
one of the given numbers which has the least number of signif¬ 
icant figures. 



FUNCTIONAL RELATIONSHIP 


69 


EXERCISE 39 

1. If U = 2 tR{R + H), find V when R = 7, H = 13, and 

TT = 

2. By the formula T = mg — mf, find T when m = 250, 
g = 32.2 and / = 75. 

3. By the same formula, find m, when T = — 35,600, 
g = 32.2, and / = 50. 

4. If S = — - % when r = — 2, I = 180, and a = 

r — i 

- 20 . 


5. By the same formula, find r when 8 — 3, a = 9 and 
I = 125. 

6. s = at — ^ gt‘^ is an important formula in physics. 

(а) Find s when a = 350, t — 4:, and g = 32.16. 

(б) Find a when s = 543, g = 32.16, and t = 4. 

(c) If t and g have fixed values, how does s change when 
a decreases? 

7. 8 = P(1 + rY is an important formula in the mathe¬ 
matics of finance. 

(a) Find 8 when P = $“5000, r = .06 and n ■= 4. (Note 
1.004 = 1 26.) 

(5) Find P when 8 = 12445, r = .05 and n = 10. (Note 
1.05^0 = 1.63.) 

(c) If r and n have fixed values, how does 8 change when 
P increases? 

8. A = is the formula for the area of a triangle on a 

180 


spherical surface of radius r. 

(a) Find A when r = 10, and E = 126°. (Use tt = 3.14.) 
Ih) Find E when A = 188 and r = 12. (Use x = 3.14.) 


9. h = K{1 4- t) is a formula from the subject of heat in 

Jdl O 


physics. Find h when A = 91 and t = 63. 




70 


ALGEBRA 


87. Finding the value of any number in certain formulas 
when the values of all the others are given. 

Recall that each variable in a formula is a function of all 
the others. (See § 78.) We are now prepared to find the value 
of any variable in certain formulas, when the others are given. 

Example. If 8 = ^ )2 a + {n - l)d\, find d when 8 = 392, 
a = 2 and n = 16. 

Solution. 1. Substituting, 392 ,= -^/{4 + 15 d\ 

2. 392 = 8{4 + 15 d}, or 392 = 32 + 120 d 

3. .•. 360 = 120 d, or d = 3 

EXERCISE 40. REMEDIAL PRACTICE 

In each of the following tables, find the missing value in each 
line, corresponding to the other values given in that line. 


V = ^(a + 5 + 4 m) 

V 

h 

a 

b 

m 

48 

12 

5 

3 


123 

18 

10 


6 

49 


10 

8 

6 


S = |{2a + («. - i)d\ 

8 

n 

a 

d 

- 99 

18 

3 



50 

- 5 

+ 3 

442 

13 


5 


r = 2 7rK(B + H) 

T 

TT 

R 

H 

1760 

3| 

14 



3.14 

10 

15 

954.56 

3.14 

8 



T = mg ~ mf 

T 

m 

9 

/ 

3400 

200 

32 


4270 


32 

20 

4270 


32 

25 





































FUNCTIONAL RELATIONSHIP 


71 


88. Making a new formula from an old one. 

Sometimes it is not as important to have definite values for 
a particular variable in a formula as it is to have a definite 
expression of that variable in terms of all the others. 

Example. Solve A = p + prt, for p. 

Solution. 1. A = p + prt, or p prt = A 

2. Factoring out the coefficient of p 

p(l A rt) = A 

3. = 

Now we have a direct expression of p in terms of A, r, and t. From it 
we can tell more easily the effect on p of changes m A, r, or t. 

Thus, if r and t are constant, p increases when A increases. If A 
is constant, p decreases when r, or t, or both increase. 


1 . 

2 . 

3. 

4. 

5. 

6 . 

7. 

8 . 

9. 

10 . 

11 . 

12 . 

13. 


EXERCISE 41 

Solve A = p + prt: (a) for r; (b) for t. 

Solve A = i h{l + c): (a) for h; (b) for c. 

Solve T = mg — mf: (a) for m; (6) for g. 

Solve E = ^ h(b -h c -t- 4 m): (a) for h, (b') for m. 
Solve *S = f w(a + /): (a) for n; (b) for a. 

Solve C = (“) 

Solve F = 2 TrR{R + H) for H. 

Solve S = -r: (a) for r; (b) for 1. 

r — t 

Solve V = -tfLl— '• ^ 

M m 

Solve s = at - i gt^ for a. 

Solve h = K{1 + for K. 

Solve S = p(l + r)^ for p. 

Solve T = 2 Ih 2 wh + Iw: (a) for h; (6) for 1. 






72 


ALGEBRA 


89. Some formulas are made by writing a general solution 
for a problem. 

Example. If an agent receives $2.00 daily and, in addition, 
a commission of 15% on his sales, then on a day when he sells 
$40.00 worth of goods, he receives $2.00 + .15 X 40, or $8.00. 

We represent his daily income by a formula as follows: 

1. Let I = his daily income. 

2. Let D = the no. of dollars worth of goods sold. 

3. .15 D = his commission. 

4. 7 = 2 + .15 D. 

Note. Clearly, the value of I depends on the value of D; as 7) 
increases I increases. 


EXERCISE 42 

1 . (a) John earns $25 per week and saves $4; how much 
will he have at the end of one year? 

(6) Write the formula for the amount John will have at the 
end of 3 years if he earns E dollars per month and spends s 
dollars. 

2. Write the formula for the amount Will earns in n hours 
at p cents per hour. 

3. Write the formula for the total weight (JV) of a bus 
weighing 5000 lb. if it is loaded with n people having an average 
weight of 140 lb. 

4. The parcel post rate for a package going to the third 
zone is 8^ for the first pound, and 2^ more, for each additional 
pound. 

(a) What is the postage for a package weighing 4 pounds? 

(b) Write the formula for the postage (p) for a package 
weighing n pounds. 

(c) By the formula, find p when n = 7. 

(d) By the formula, find n when p = 30, 

(e) If n increases then p __ 



FUNCTIONAL RELATIONSHIP 


73 


6. Write the formula for the number of tons of coal in a bin 
whose dimensions are L, W, and H feet respectively, if one 
ton occupies 38 cu. ft. 

6. The taxi rate in one community is 25 for the first one- 
half mile, and additional for every additional one-half mile. 

(а) What is the charge for a ride of miles? 

(б) Write the formula for the charge (c) for a ride of n miles. 

(c) By your formula, find c when is 6. 

7. The rate for a person to person long distance telephone 
call from Madison, Wis. to Chicago is $1.00 for the first three 
minutes, plus 30^ more for each additional minute. 

(а) What is the charge for such a conversation lasting 6 
minutes? 

(б) Write the formula for the charge (c) for a conversation 
lasting n minutes. 

(c) By your formula, how long may one talk for the sum of 
$2.50? 

id) When n increases then-. 

8. The rate for sending a telegram from Madison, Wis. to 
Minneapolis is 48^ for the first 10 words, plus 3^^ for each 
additional word. Write the formula for the cost (c) of a tele¬ 
gram containing n words. 

9. Teachers in a certain community are paid $1250 per 
year, plus $50 for each additional year of service. 

(а) Write the formula for the salary (s) of a teacher who is 
teaching her nTh year. 

(б) What is the salary for a teacher who is teaching her fifth 
year in the community? 

(c) If n increases then s _ 

10. Write the formula for the average of the marks of a class 
if a pupils receive the mark 90, h the mark 80, c the mark 70 and 
d the mark 60. 

Note. Additional examples appear on p. 242. 





74 


ALGEBRA 


EXERCISE 43. CHAPTER MASTERY-TEST 

1. One number which_J-one or more other 

_is said to be a-of them. 

2. A functional relation between two or more numbers can 

be expressed in-ways; namely, (give them). 

3. In the functional relation V = h(b + c + 4 m), if h is 

constant then F is a function of-; then if h, c, and m 

all increase, V _If h, c, and m are constant, then V 

is a function of h] then if h is doubled V is- 

4. {a) The formula A = p -f- prt expresses the - 

_between_ 

(6) Simplify the formula when r = .05 and t = 4. 

(c) Using the simplified formula, make a table of correspond¬ 
ing values of A and p for p = 0, 100, 200, etc. to 1000 inclusive. 

(d) Represent the table of part (c) by a graph. 

5. {a) Express in words the functional relation = 4 ttA, 
in which S represents “area of sphere” and r the radius. 

(6) In this formula, TT is a_; and r are_ 

(c) If r is the independent variable, then S is the_ 


EXERCISE 44. CUMULATIVE REVIEW 

1. Divide 4- by a: — y, finding the quotient and re¬ 
mainder. Check your result by using the rule connecting 
dividend, divisor, quotient, and remainder. 

3a: — 6 x‘^ — 4 -50: + 10 


2. Simplify 


3. Combine 


4x4-4 

t 


3 X — 3 
^2 


X^ — 1 


1 - 


5 -f- ^ 
1 


4. The quantity 


(1 + 


is from the mathematics of 


finance. Find its value when i = .05- 























VI. SYSTEMS OF FIRST DEGREE EQUATIONS 

90. Can you pass the following test? If you cannot you need 
the remedial instruction and practice on pages 76 to 84. If 
you can, you will not require much time for these pages and 
can go on to the new subject matter on pages 85 to 93. 

DIAGNOSTIC TEST 9 

1. An equation like 3 a: + 2 y = 12 has_solutions. 

One such is_ 

2. Two equations like Z x 2 y = 12 and x — y = — 1 

form a _ of equations. Usually they have - 

common solution. When they do have, they are called- 

equations. 

3. The graph of 3 x + 2 y = 12 is - 

It can be obtained by locating the points 

A\ X = ; y = ; and B: x = ; y = ; whose co¬ 
ordinates are-of the equation. 

4. The common solution of 3 a: + 2 ^ = 12 and x — y = — 1 

can be obtained_or by - methods of 

elimination. 

6 The common solution oiS x 2 y = 12 and x — y = — 1 
obtained graphically is- 

6. The common solution obtained by the-method 

of elimination is -- 

7. Determine graphically whether ^ + y = 5 and 3 a: + 3 y 
= 6 have a common solution. Such equations are called 
_equations. 

8. Determine graphically the common solution oi x y 

= 5 and 3 a; + 3 ^ = 15. These are-equations. 

75 
















76 


ALGEBRA 


Remedial Instruction and Practice, Pages 76 to 84 


91. Consider the equation 3 x — 4 z/ = 12. 

It expresses a functional relation between x and y, because: 

(a) for every value of x, there is a value of y. 

Thus, when a: = — 4, ?/ = — 6. .’. ?/ is a function of x. 

(b) for every value of y, there is a value of x. 

Thus, when y = S, x = S. .'. a: is a function of y. 

92. A solution of an equation having two variables consists 
of a value of one variable and the corresponding value of the 
other variable, which together satisfy the equation. 

By §91, a, or h, there is an infinite number of solutions of 
an equation like 3a: — 4?/ = 12. 


93. We can represent 3 x — 4 z/ = 12, graphically. 

Y 



The graph is a straight line. 


Observe point A on the graph. Its coordinates are x = 5; 
y = f. These satisfy the equation, for 3 • 5 — 4(f) = 15 — 3, 
or 12. 


Observe point B, not on the graph. Its coordinates are 
X = 2.5; y = — 2.5. These do not satisfy the equation, for 
3(2.5) - 4(-2.5) = 7.5 + 10, or + 17.5. 






































































SYSTEMS OF FIRST DEGREE EQUATIONS 77 


94. An equation like 3a: — 4i/ = 12 is a rational and integral 
equation of the first degree having two variables. 

It is rational since neither x nor y is under a radical sign. 

It is integral since neither x nor y is in a denominator. 

It is of the first degree since the sum of the exponents of x 
and y is one or less in each term of the equation. 

95. The graph of a rational and integral equation of the first 
degree haring two variables is always a straight line. 

Also, every straight line has as equation such a first degree 
rational and integral equation. 

Since every point on the y-axis has 0 as abscissa, the equa¬ 
tion of the y-axis is x = 0. 

Since every point on the x-axis has 0 as ordinate, the equation 
of the x-axis is y = 0. 

Since every point on the line parallel to the y-axis through 
the point x — 2, y = 0, has abscissa 2, the equation of this 
line is X = 2. Similarly for other lines parallel to the y-axis 
and lines parallel to the x-axis. 

Rule. To obtain the graph of a linear equation. 

1. Find three common solutions of the equation. 

2. Locate the points whose coordinates are these solutions, and 
draw the line connecting them. 

If it is a straight line, the work is done correctly. 

3. The x-intercept is the distance from the origin to the point 
where the line crosses the x-axis. It is found hy letting y = 0. 

Similarly the y-intercept is found hy letting x = 0. 

EXERCISE 45 

Draw the graphs of the following equations. 


1. 

y = X 

6. 

* + 2/ = 8 

9. 

5 X + 2 y = 20 

2. 

y = 

6. 

a; + 2 2 / = 4 

10. 

3 X = 5 — 2 y 

3. 

II 

7. 

Zx — y = & 

11. 

1 

II 

4. 

X = + 4 

8. 

Zx-2y= 12 

12. 

X = — 3 


78 


ALGEBRA 


96. Two linear equations studied graphically. 

Example. What must be the length and the width of a rec¬ 
tangle to have a perimeter of 36 and a length 3 more than twice 
the width? 


Solution. 1. Let I = the length and w = the width. 

2. •*. 2 Z + 2 ly = 36, or Z + ly = 18. 

and I — 2 w = S. 

3. Draw the graphs of the equations. 



Z + ly = 18 

Z 

— 2 w = 3 

Point A: 

when Z = 10, ly = 8 

Point D : 

when Z = 3, ly = 0 

“ B: 

“ Z = 6, w; = 12 

“ E: 

“ Z = 9, ^y = 3 

“ C: 

» Z = 2, -y; = 16 

“ F: 

“ Z = 15, ly = 6 


4. The graphs are at the right. 

Line ABC represents graphically the 

relation between the length and width 
of a rectangle with perimeter 36. 

Line DEF represents graphically the 
relation between the length and width 
of a rectangle when the length exceeds 
twice the width by 3. 

5. Observe point N. Its coordinates 
are 13 and 5. Since it is on both lines, 
its coordinates satisfy both conditions. 

That is Z + ly = 18, or 2 Z + 2 ly = 36, 
as required. Also Z — 2 ly = 3, as required. 

Remarks. 1. Since there is only one point of intersection of two 
straight lines, then there is only one common solution for the two 
equations, and only one solution for the problem. 

2. Each equation has solutions which are not solutions of the other. 
Thus the coordinates of point B satisfy the equation Z + ly = 18 but 
do not satisfy I — 2 w = 3. 



97. Two equations having two variables are independent if 
each has solutions which are not solutions of the other. 

Two independent equations which do have common solutions 
are called simultaneous equations. 










































































SYSTEMS OF FIRST DEGREE EQUATIONS 79 

Two independent equations which do not have a common 
solution are inconsistent equations. 

Two equations having two variables are dependent if every 
solution of each is a solution of the other. 

98. Two simultaneous linear equations having two variables 
have only one common solution. 

Rule. To determine graphically the common solution of two 
simultaneous linear equations having two variables: 

1. Draw the graphs of the two equations upon one set of 
axes. 

2. Determine the coordinates of the point which is on both 
graphs. This is the common solution. Check the common solu¬ 
tion by substituting it in both equations. 

EXERCISE 46. REMEDIAL PRACTICE 

Study the following systems of equations graphically. If 
they are simultaneous, give their common solution. 



80 


ALGEBRA 


99. The addition or subtraction method of eliminating one 
variable. 


Example. Solve the system 


b X y = — 9 
3a; — 42/ = — 17 


( 1 ) 

( 2 ) 


Solution. 1. The coefficients of y are unlike. We can make them 
alike except for sign by multiplying equation (1) by 4 and equation 
(2) by 3. Then, if we add, the y terms will drop out. 

M 4 ( 1 ) * Then 2Q x + 12 y = - 36 (3) 

2. M3(2) 9 a; - 12 2 / = - 51 (4) 

3. (3) + (4) 29 X = - 87 (5) 

4. D 29 X '= — 3 

5. Substitute — 3 for a; in ( 1 ) — 15 + 3 2 / = — 9 

6. 3 2 / = 6, or 2 / = 2 

7. The solution is x = — Z, y = 2 

Check. In (1): Does 5(- 3) + 3(2) = - 9? 

Does —15 + 6= — 9? Yes. 

In (2): Does 3(- 3) - 4(2) = - 17? 

Does — 9 — 8 = — 17? Yes. 


Note 1. Sometimes one variable has the same coefficient in both 
equations. In this case subtract instead of adding as in Step 3. 


Rule. To eliminate one variable from two equations by 
addition or subtraction. 

1. If necessary, multiply both equations by such numbers as 
will make the coefficients of one of the variables of the same abso¬ 
lute value. 

2. If the coefficients have the same sign, subtract one equation 
from the other; if they have opposite signs, add the equations. 

3. Solve the equation found in Step 2 for the other variable. 

4. Substitute the value of the variable found in Step 3 in any 
equation containing both variables, and solve for the remaining 
variable. 

5. Check by substitution in the original equations. 

Note. Remember that this common solution consists of the co¬ 
ordinates of the point on the two graphs. 

* Note. M4(1) means “multiply both sides of (1) by 4.” 


SYSTEMS OF FIRST DEGREE EQUATIONS 81 


EXERCISE 47. REMEDIAL PRACTICE 


Solve the following systems of equations by addition or sub¬ 
traction. (If difficulty is experienced in obtaining a solution, 
determine graphically whether the equations are inconsistent 
or dependent.) 


8 X + 5 y = 5 
3 X - 2 2/ = 29 
7 r — Q s = 63 

9 r + 2 5 = 13 

3 c + 7 d = - 23 
5c + 4d = -23 
z - 6y = 2 

2 X - 12 1/ = 21 

5 X Sy = —9 
3x-4i/ = -17 
5 a “h 4 6 = 22 

3 4- 6 = 9 


7. 

8 . 

9. 

10 . 

11 . 

12 . 


6x + 2z = - 3 
5x — 3^! = — 6 
8 m 9 n = 3 
8 m — 9 n = 77 
Ss+7t = 4: 

7 s + 8t= 26 
7c - 2d = 31 

4 c - 3 d = 27 

5 m + 4 = 22 

3 m + n = 9 

r — 6 s = — 10 
2r - 7s = - 15 


In Ex. 13 to Ex. 15 simplify 
by clearing of fractions. 


13. 


14. 


2 X Sjy _ _ 7 
4 “ 2 


5 2 

6-5 


7 

a + 3 


= 11 


10 a - 
86 - 

5 _ 

c - 2 3 - 2d 

1 8 


= - 17 
7 


2c+5 7-3d 


= 0 
= 0 


In Ex. 16 M 3 (1); then 
M 5 ( 2 ) ; then add. 


16. 


17. 


18. 


2p + -= - 11 

q 

A 3 21 

3-1 = 9 

r s 

- 1 

r s 

12-2 = 2 

X y 

§ _ = -1 
X y 


15. 














82 


ALGEBRA 


100. The substitution method of eliminating one variable. 

This method can safely be omitted from a minimum course. 


Example. Solve the system 


11 X 
4 X 


5y = 4:. 
3y = 5. 


Solution. 1. First solve equation (1) for x, as follows: 

4 + 5y 
11 


Aey 11 X = 4 5 y; Dn x = 

[4 -f- 5 y] 

5. Substitute this value of a; in (2). 4 |—— | — 


( 1 ) 

( 2 ) 

(3) 

(4) 


3. 


16 + 20 ?/ 
11 


- 32 / = 5 


4. Mil 16 + 20 2/ - 33 2/ = 55 

5. Si6 ••• - 13 2 / = 39, or 2 / = - 3 

6. Substitute — 3 for ^ in (2). .*. 4 a: + 9 = 5 

7. .'. 4 a; = — 4, or x = — 1. 

8. .’. the common solution is: x = — 1; ?/ = — 3. 

Check by substituting in equations (1) and (2). 

Note 1. You can solve either equation for one variable in terms 
of the other. Naturally, you select the one which can be solved most 
easily. 

Note 2. Try to do all of Step 1 mentally. 

32/4-5 „ llx-4 

4 


Thus from (2) x = 
From (2) 


y = 


From (1) y = 
4 X — 5 


Rule. To ehminate one variable from two equations by 
substitution. 

1. Solve one equation for one variable in terms of the other. 

2. Substitute for this variable in the other equation the value 
found for it in Step 1 . 

3. Solve the equation found in Step 2 for the second variable. 

4. Substitute the value of the second variable, obtained in Step 3, 
in any equation containing both variables and solve for the first 
variable. 

5. Check by substitution in the original equations. 








SYSTEMS OF FIRST DEGREE EQUATIONS 83 


EXERCISE 48. REMEDIAL PRACTICE 
Solve the following systems by substitution. 

(If difficulty is experienced in securing a common solution, 
determine graphically whether the equations are inconsistent 
or dependent.) 


3a: + y = 11 
bx — y = 13 
3a:H- y = Z 
4a: + 2y = - 
2a: + 3?/ = 4 
a: - 4?/ = - 


4. 


9 


3 a: + 4?/ = 3 
2 a: — y = 13 
a: + 5 2 / = 1 
2 a: + 7 2 / = 1 
4x + 3 2/ = 2 
2x - 2/ = f 


First reduce each equation of the following examples to the 
form ax hy = c, which is called the normal form. 

2 X - 12/ = 13 


7. 


8 . 


13. 


14. 


16. 



10 . 


11 . 


12 . 


16. 


17. 


18. 


4 ^ ^ 5 

2 a + 18 = 6 
ix + f 2/ = 2 


ix - f y = 6 
f 2/ + = 0 

r t r — t 


2 

3 

r + / , 

r — t 

3 + 

4 

X - y 

2x + y 


2x-3 22/+ 13 


3 

a: + 2 2/ _ 
2 

r + ^ — 9 
2 

r — s 


= 0 
n 

4 


2s 

6 


3 

r s 



























84 


ALGEBRA 


101. Problems about a number and its digits. 

Integers are written by means of the digits 0, 1, 2, 3, 4, 5, 6, 
7, 8, 9. 

Thus, 38 is a 2 digit number. 3, its tens’ digit, represents 30 units, 
8 represents units; altogether, thirty-eight units. The sum of the 
digits is 11; their product is 24. 

If t is the tens’ digit of a number and u is its units’ digit, the 
number contains (10 ^ + u) units. 

When the digits are reversed a new number is formed. 

Thus reversing the digits of 52 we get 25. 

Observe that 52 = (5 X 10 + 2), or fifty-two units. 

25 = (2 X 10 + 5), or twenty-five units. 

EXERCISE 49. REMEDIAL PRACTICE 

1 . If a certain number of two digits be divided by the sum of 
its digits, the quotient is 4 and the remainder is 3. If the 
digits be reversed, the sum of the resulting number and 23 is 
twice the given number. Find the number. 

Solution. 1. Let t = the tens’ digit and u = the units’ digit. 

2. .•. 10 f + w = the given number, 

and 10 u t — the number with digits reversed. 

3. lot + u = 4d + w) + 3. (1) 

4. lOu + t + 23 = 2(10 1 + u). (2) 

(Complete the solution.) 

2. The tens’ digit of a number of two digits exceeds its units’ 
digit by 4. If the digits be reversed, the new number is 6 more 
than one half the old number. Find the number. 

3. The sum of the two digits of a number is 16; if 18 be 
subtracted from the original number, the remainder equals the 
number obtained by reversing the digits. Find the number. 

4. If a certain number be divided by the sum of its two digits, 
the quotient is 7 and the remainder is 6; if the digits be reversed, 
the sum of the new number and twice the old number is 204. 
Find the number. 


SYSTEMS OF FIRST DEGREE EQUATIONS 85 


102. Problems about opposing forces. 

As you know a river has a current which exerts a force down¬ 
stream. Rowing a boat downstream is therefore easier than 
rowing upstream. 

If the current flows at the rate of 2 miles per hour, then, 
theoretically, any one rowing downstream at the rate of 4 miles 
per hour will proceed at the rate of (4 + 2) or 6 miles per hour; 
upstream, the rate will be (4 — 2) or 2 miles per hour. 

EXERCISE 50 

1 . If a boat is rowed downstream 10 miles in two hours and 
the same distance upstream in 3| hours And the rate of rowing 
in still water and the rate of the current. 

Solution. 1. Let r = the rate of rowing in still water 
“ c = the rate of the current 
(r + c) = the rate downstream 

2. and (r — c) = the rate upstream 

3. .-.^ = 2 (Since = time) 

r + c rate 

and —- - = 
r — c ^ 

(Complete the solution.) 

2 . A crew can row 18 miles downstream in 2 hours, and back 
again in 3 hours. What is the rate of the crew in still water 
and the rate of the current? 

3. An airplane traveled 200 miles with wind of a certain 
velocity in 4 hours; it returned against a wind of the same 
velocity in 6f hours. What was the rate of the wind and of 
the airplane? 

4. A motor boat, which can run at the rate of r miles per hour 
in still water, went downstream a certain distance in n hours; 
it took m hours to return. 

(a) Find the distance and the rate of the current. 

(b) The distance is a function of- 






86 


ALGEBRA 


103. Problems about reciprocals. (Optional) 

(a) The reciprocal of 3 is 1 3, or of f is 1 f, or f. 

(b) If a piece of work can be done in 5 days, then one fifth of 
it can be done in 1 day, and n fifths of it in days. 

If a piece of work can be done in x days, then 1 -i- x part of it 
can be done in 1 day, 3 x part in 3 days, and n x part in 
n days. 

EXERCISE 51. {Optional) 

1. A piece of work can be done by A and B working together 
in 10 days. After working together for 7 days, A stops, and B 
completes it in 9 days. How long would it take each to do the 
work alone? 


Solution. 1. Let a = the no. of days it would take A alone 
and 6 = the no. of days it would take B alone. 


Person 

Time 

Alone 

Amount 

IN 1 Da. 

Amount 

IN 10 Da. 

In 7 Da. 

In 9 Da. 



1 

10 

7 


A 

a Da. 




— 



a 

a 

a 


B 

b Da. 

1 

15 

7 

9 



h 

b 

b 

b 


3. 


10 , 10 ah 10 , 10 , 

a b ah a b 

,7,7,9 ab 7 , 16 , 

and — 1" r “b r ~ or ——= 1 , 
a b b ah’ a b 

(Complete this solution as in Ex. 16, p. 81.) 


2. The sum of the reciprocals of certain two numbers is f. 
Twice the reciprocal of the larger increased by three times the 
reciprocal of the smaller is 7. What are the numbers? 

3. A and B, working together, did a certain piece of work 
in 6 days. At another time, they did the same amount when 
A worked 3 days and B 10 days. How many days would it 
take for either to do the work alone? 










SYSTEMS OF FIRST DEGREE EQUATIONS 87 


104. Solution of systems of literal equations. {Optional) 
Problems of this type do not cause any special difficulty. 


Simply remember and use this fact: 

ax hx — cx = (a + 6 — c)x. 


Example. Solve the system U “ + = 4 + 6^ 

^3!^ ^ y — ^ CL U 

(1) 

(2) 

Solution. 1. Mb (2) 

hx — 2 by = 2 ab — 

(3) 

2. Add (1) and (3) 

2 ax + bx = 4: 2 ab 

(4) 


or x{2 a + b) = 2 a{2 a + 5) 

(5) 

3. D(2a+6) 

:. X = 2a 


4. Substitute in (1) 

4 a2 + 2 &2/ = 4 a2 + 62 



5. ••• 2 by = OT y = ^ 

Check. Substitute in (2); Does 2 a — 2 = 2 a — 6? Yes. 


EXERCISE 52. {Optional) 



2 ax -\r y = b 

6. 

\ ax — by = ai^ b^ 

ax — 2 y = c 

\ ax by = a^ b^ 

2. « 

[ mx + ny = rn} S mn 

1 

2bx -\- ay = a + 6 

( 2 mx — 3 Tiy = 2 — 4 mn 

2 abx + aby = a^ b‘^ 

1 

o 

II 

1 

8. < 

f mx -\- ay = w? n 

[ abx — ahy = ai^ — ¥ 

( X + amy = m + mn 

1 

\ cx -\r dy = c + d 

9. ^ 

\ acx = by 

[ cdx + cdy = 

[ obex + aby = (a + by 

5. < 

\ ax — by = c 
[bx ay = d 

10. ^ 

\ sy = mx 
( sx + rsy = t 


11. Find two numbers whose sum is s and difference is d. 

12. Separate N into two parts whose quotient shall be a and 
remainder h. (The results are formulas for the two parts.) 

13. (a) The perimeter of a rectangle is P. Its base exceeds 
its altitude by d. Find the base and the altitude. 

(6) Your result shows that 6 is a function of- 



88 


ALGEBRA 


105. Systems of equations having three or more variables. 

When there are two variables, then two equations are neces¬ 
sary to obtain a single common solution. Similarly when there 
are three variables, then three equations are necessary to obtain 
a single common solution; four equations, if there are four 
variables; etc. 

However, just as two equations having two unknowns may be 
dependent, or be inconsistent, so a system of three equations 


may not have 

a single common solution. The complete 

dis- 

cussion of these facts appears in a later course in algebra. 


Example. Solve the system of equations: 



ri2m — 4?i+ p = 3 

(1) 


j m — n — 2 p = — 1 

(2) 


\ b m — 2 n =0 

(3) 

Solution. 1. 

M2(1). 24 m — 8n + 2p = 6 

(4) 

2. (2) + (4) 

25 m — 9 n = 5 

(5) 

3. M5(3) 

25 m — 10 w = 0 

(6) 

4. (5) - (6) 

n = 5 



5. Substitute 5 for n in (3). 5 m — 10 = 0, or m = 2. 

6. Substitute 5 for n and 2 for m in (2). 


2 - 5 - 2p = - 1 

7. — 2 p = 2, or p = — 1 

8. .'. w? = 2, n = 5, p = — 1 

Check. The solution satisfies each of the three given equations. 

Rule. To solve a system containing three variables: 

1. From two equations, say the and the 2nd, eliminate one 
of the variables; mark the resulting equation, equation (4). 

2. From another pair of equations, say the 1st and Srd, elimi¬ 
nate the same variable; mark the resulting equation, equation (5), 

3. Solve the system consisting of equations (4) and (5). 

4. Determine the value of the third variable, by substituting 
in one of the given equations the values found in Step 3. 

5. Check by substitution in the given equations. 


SYSTEMS OF FIRST DEGREE EQUATIONS 89 


EXERCISE 53 


Solve and check the following systems of equations: 


1 . 


2 . 


4. 


11 . 


a — 26+ c = 5 
2a+ 6— c=— 1 

3a + 36 — 2c = —4 
3a:+ y — 2=11 

a: + 3y— 2 = 13 

X + I/ — 32 = 11 
2r+35— t = — 2 
4:r - Ss 2t = 9 
6f — 05+3^ = 13 
in-\-6n-\-3p = 8 
3m + 4w = — 3 

5 m + 7 n = 1 

3 + — B 2 C = — 

6 + + 3.6 + 2f7 = — 
3 + + R-4C = 5 


10 . 


{2x-\-8y- 2 = 

\ 3 X — 8 z = — 1 
[52/+ 72= -1 
f2x + y — z=0 
+ x + z = 4)/ 

{y = X \ 

(3x + y — z = 2 
\y=l-2x 
[32= - 2y 
r2x — 2/ + 42 = 0 
\ X — y + 3z=0 
[3x-22 = f 
fx-\-2y — z=7 
^ y 3x = z 
[2x y = -1 


2 1 

3 



f 1 

1 



7 


— - 

-= a 

X y 

Z 



X 

y 

1 2 

1 



1 

1 

-+ - - 


10 

12. < 


- = 6 

X y 

2 



X 

2 


3 _ 3 ^ 2 ^ 

X y z 


i + i = . 

y 2 


- 1 


13. The perimeter of a certain triangle is 100 inches. The sum 
of the first and second sides exceeds the third side by one half 
the third side. The sum of the second and third sides equals 
three sides the first side. What are the lengths of the sides? 

14. The sum of the digits of a certain number of three digits 
is 11. If the number be divided by the sum of its hundreds’ 
and its units’ digits, the quotient is 20 and the remainder is 6. 
The units’ digit exceeds the sum of the hundreds’ and tens’ 
digits by 1. Find the number. 






90 


ALGEBRA 


106. Solving a set of equations by some formulas. {Optional) 


Consider the set of equations 


ax -\- by = c, 
dx ey = /. 


If solved as in § 103, 1 
you will find that: J 


ce — bf _ af — cd 
ae — bd’ ^ ae — bd 


Some mathematician saw that ae — bd can be obtained by 
a b 


arranging a, b, d, and e thus 


d e 


and taking the product of the 


downward diagonal minus the product of the upward diagonal. 


a b 
d e 


is defined as being ae — bd. It is called a determinant. 


Similarly 


c b 

f e 


hf; 


a c 

df 


= af — cd. 


Therefore x = 


c b 


a c 

f e 


df 

a b 

y - 

a b 

d e 


d e 


The advantage of these formulas is obvious from the rule. 
Rule. To solve a set of two equations by determinants: 

1. Arrange the equations in the form: J ^2/ 

a b 


2. There is a common solution if 


d 


dx+ ey = f. 
is not zero. 


3. The value of x is a fraction; its denominator is the deter¬ 
minant formed by the coefficients of x and y; its numerator is 
the determinant obtained by replacing the coefficients of x in the 
denominator determinant by the corresponding absolute terms. 

4. The value of y is a fraction with the same denominator as x; 
its numerator is the determinant obtained by replacing the coeffi¬ 
cients of y in the denominator determinant by the absolute terms. 




















SYSTEMS OF FIRST DEGREE EQUATIONS 91 


Example 1. Solve 


Solution. 1. 


1 “ 

16-51 

5 71 

1 

2-51 

1 

3 7l 


2-161 


3 5 


2 - 5 


3 71 


2x - 5y = - 
Sx + 7 y 6 


16 


by determinants. 


7(- 16) - 5(- 5) 
2 • 7 - 3(- 5) 


- 112 + 25 
14 + 15 


= - 3 


2 • 5 - 3(- 16) 10 + 48 58 
2 • 7 - 3(- 5) 14 + 15 29 


Example 2. Consider the system 


2 X + 3 2/ = 6 
4 X + 6 2/ = 24 


( 1 ) 

( 2 ) 



Example 3. Consider the system 


2 X + 3 2/ = 6 
4 X + 6 2/ = 12 


1 6 31 


|2 61 

|l2 6 

36 - 36 0. 

4 I 2 I 

1 2 31 
1 4 6 

12-12 0’ ^ 

12 31 
14 6l 


These two equations are dependent. 

Rule. If the denominator determinant is zero, the equations 
are either inconsistent or dependent. They are dependent if, 
besides, both numerators are zero; they are inconsistent if at least 
one numerator is not zero. 






































































92 


ALGEBRA 


EXERCISE 54. (PptionaV) 


107. Solve by determinants the following systems. If they 
are not simultaneous, tell whether they are inconsistent or 
dependent. 


1 . 

2 . 

3. 

4. 

6. 


a -\- h = b 
a — 6 = 9 
x-\-2y = 12 
2tx — y = I 
3x - 2y = -12 
4:x + 2y = - 2 
2a: + y = 4 
Qx-^Sy = 12 
5 a: + 2y = 10 
X - y = 9 


6 . 

7. 

8 . 

9. 

10 . 


3 a — 2 6 = 6 
6 a — 4 6 = 18 
5 X — 6 y = 8 
9 X — 4 y = — 6 
8 p + b q — 5 
3 p — 2 q = 29 

2 X — 3 y = — 14 

3 X + 7 y = 48 

r — 0 5 = 2 

2 r - 12 5 = 10 


108. Determinants are especially useful in solving sets of 
equations in three or more variables. {Optional) 

To solve a set of three equations with three variables, de¬ 
terminants of three rows and three columns are used. 


is a determinant of the third order. 

It is defined to have the value 
aih^cz + a^h^ci + a^hic^. — — a^hic^ — aih^C2 

This value may be obtained from the determinant as fol¬ 
lows: From the sum of the products of the downward to right 
diagonals subtract the products of the upward to right diagonals. 


ai 02 as 
bi 62 63 

Cl C2 C3 


Example. 

1 5 2 

4 7 3 =l-7-6+5-3-2+2(-3)-4 

2 -3 6 -2-7-2-4-5-6-l-3(-3) 

= 42 + 30 - 24 - 28 - 120 + 9, or - 91 
= - 91. 






SYSTEMS OF FIRST DEGREE EQUATIONS 93 


EXERCISE 55 

Find the value of: 


1. 

1 2 3 

2 12 

2. 

2 4 6 

3-23 

3. 

2 2 3 

- 2 - 4 - 11 


3 3 1 


1 5 4 


5-6 2 


4. Solve by determinants the equations: 

f S X + y — 2 = 14 

j X S y — z=16 

[ X y — Sz = — 10 

Solution. A rule similar to that of § 106 applies for linear equations 
with more than two unknowns. Namely: 


y = 


14 1 - 1 
16 3 - 1 
- 10 1 - 3 

- 126 + 10 - 16 - 30 + 48 + 

3 1-1 

-27-1-1 + 3 + 3 + 3 

13-1 



11-3 



3 14-1 


3 1 14 

1 16 1 


1 3 16 

1-10-3 

1 

1 

to 

0 

1 

55 

1 

1 1 - 10 

3 1-1 

- 20 

3 1 - 1 

1 3-1 


13 - 1 

1 1-3 


11 - 3 


- 100 ^ 



Check. The solution checks when substituted in the three equations. 
Note. The equations must be arranged first in the form ax + by 
cz = d. Thus the equation 2 x — S z = 7 would be written 2x-{- 
0 y — S z = 7. 


Solve the following equations by determinants: 


f X y — z = 24: 
5. j4a:d-32/ — 2=61 
[6a: — 5y — 2 = 11 


(4:X-Sy=l 
7. j 4 y - 3 2 = - 15 
[ 4 2 - 3 a: = 10 


f 5 X — y + 42=—5 
B. \ 3a:+5y+62 = -20 
[ a:+3y — 82 = — 27 


r9a: + 52 = - 7 
< Sx-\- 5y = 1 
[9?/ + 32 = 2 


8. 


























94 


ALGEBRA 


EXERCISE 56 

1. If the numerator of a certain fraction be trebled, and the 
denominator be increased by 7, the value of the resulting frac¬ 
tion is J. If the denominator be doubled and the numerator 
be increased by 3, the value of the resulting fraction is 
Find the fraction. 

2. The perimeter of a certain triangle is 100 inches. The 
sum of the first and second sides exceeds the third side by 
12. The sum of the second and third sides equals four times 
the first side. What are the lengths of the sides of the 
triangle? 

3. The sum of the reciprocals of certain three numbers is 
— The reciprocal of the first exceeds that of the second by 
J. The reciprocal of the first, diminished by twice that of the 
third, is f. What are the numbers? 

4. If a rectangular lot were 6 feet longer and 5 feet wider 
than it is now, it would contain 839 square feet more; if it were 
4 feet longer and 7 feet wider, it would contain 879 square feet 
more. Find its length and its width. 

5. In a certain triangle, ABC, angle A exceeds angle B by 
15°. The sum of angle A and twice angle C is 150°. Find the 
size of each angle. 

6. A crew can row 15 miles downstreain in 2 hours, and 
back the same distance in 3 hours. What is the rate of the 
crew in still water, and the rate of the current? 

7. If A and B do a piece of work in 10 days, B and C in 
20 days, and A and C in 12 days, how many days would it 
take each to do it alone? 

8 . The sum of the digits of a certain number of three digits 
is 17. The tens’ digit exceeds the units’ digit by 5. If the digits 
be reversed, the old number is 38 more than three times the 
new number. Find the number. 


SYSTEMS OF FIRST DEGREE EQUATIONS 95 


109 . We are now able to obtain without any guessing the 
formula connecting two numbers when we have a table of 
related values of these numbers, — that is for certain cases. 
For other cases we must wait until later. This is one of the 
very practical uses of graphs and functional relationships in 
applied mathematics. 


Example. Assume that an experiment involving the measure¬ 
ment of two numbers K and n has produced the following table 
of related values. 

When n = \, 3, 4, 5 
then K = 10, 14, 16, 18. 

What formula, if any, connects them? 

Solution. 1. First draw the graph for 
these pairs of numbers. 

2. The points appear to fall on a straight 
line. This suggests that the formula is of 
the form 

K = an h. 

In this formula, a and h are unknown. 

3. Substitute two sets of values from the 
table in this formula. 

Thus 

when n = 1, A = 10. a + 6 = 10. 

when n = 3, A = 14. .-. 3 a + 6 = 14. ' ^ ^ ^ ^ ^ 

4. Solving this system, a == 2 and h — S. 

Hence the formula appears to be A = 2 n + 8. 

5. Does it check? Try the pair of values n = 5 and A = 18. Does 
18 = 2 X 5 + 8? Yes. 

Try n = 4 and A = 16. Does 16 = 2 X 4 + 8? Yes. 

Therefore the formula is A = 2 n + 8. 

Note. Understand, now, that the plan is to draw the graph for 
the table of related values. If the graph turns out to be a straight 
line, then we infer that one of the numbers is a first degree function 
of the other. If the graph is not a straight line, then the one variable 
is not a first degree function of the other, and we cannot solve the prob¬ 
lem at the present time. 

(Examples appear on page 96.) 










































96 


ALGEBRA 


EXERCISE 57 

If possible, derive the formula for each of the following tables 
of related variables. Follow the plan taught on page 94. 

1 . (a) The parcel post rates to the third zone are: 

weight (w) 4 lb. 7 lb. 9 lb. 12 lb. 
rate (r) 14jzf 20^ 24ff 
Find the formula connecting w and r. 

(6) From your formula give the rates for 1, 2, and 5 pounds, 
(c) From your formula tell how r increases when w increases 
1 lb. 

2. (a) The rates for the sixth zone are: 

weight (w) 5 lb. 10 lb. 15 lb. 20 lb. 

rate (r) 42^ 82|zf 1.22 1.62 

{b) From your formula, find the rates for 1, 2, and 3 pounds, 
(c) From your formula tell how r increases when w increases 

1 lb. 

3. (a) The rates for expressing a package not over 50 miles 
are: 

weight (w) 10 lb. 20 lb. 25 lb. 

rate (r) 43^ 51 55^ 

(b) From your formula find the rates for 5 lb. and 15 lb. 
Note. The rate is the same for 0 to 5 lb. inclusive; etc. 

4. An object falling freely was found to pass over the fol¬ 
lowing distances in the time given. 

Time 1 second 3 seconds 4 seconds 
Distance 16 ft. 80 ft. 112 ft. 

Find the formula connecting the time t and the distance d. 

5. After an experiment it was found that two numbers W 
and d are related as follows: 

When d = 1 2 3 4 

then W = 27 51 75 99 
Find the formula connecting them, if possible. 


SYSTEMS OF FIRST DEGREE EQUATIONS 97 


EXERCISE 58. CHAPTER MASTERY-TEST 


Solve the following systems of equations if possible and 
tell what kind of graphs they produce. If you cannot find a 
common solution, tell what kind of equations they are and what 
the graphs will be. 


1 . 

2 . 

3. 


3 a: + 5 2/ = 19 

4x — 2 / = 10 

2x + y = 5 

4 a: + 2 2/ = 10 



Solve by determinants: 


4 X - 3 2/ = 1 
3 X — 4 2/ = 6 


4. 

6 . 

6 . 


f 6 X — 2 ^ = 15 
\ 3 X — ^ = 4 

I 2 X — y = 5a— h 

l3x + 22/ = lla+26 


Sx - y 
3 

2x - y 
3 


4 X - 2 2/ - 8 
2 

5x4-32/ 


I" a — 2 h -\~ c = 0 
8. \ a — h 2 c = — 11 

[ 2 a— 64 - c=— 9 


9. An equation like — 3x4-52/ =16 expresses a_ 

relation between_and_Solved for y, y = _ 

This last equation shows that an increase of 1 unit in x produces 
_of_units in y. 

10. The functional relation y = .5x4-7 shows that a de¬ 
crease of 1 unit in x produces a_of_units in y. 

The graph is a_line which rises_spaces for each 

space to the right; it crosses the 2 /-axis at the point y = - 

and the x-axis at the point x =_ 


11. What formula connects 1 when s = 5 8 10 13 

w and 5 if: | then w = 3.5 3.8 4 4.3 

12. If a two digit number be divided by the sum of its digits 
the quotient is 5 and the remainder is 8. If the digits be re¬ 
versed, and the new number be divided by the sum of its digits, 
the quotient is 4 and the remainder is 6. What is the number? 

















98 


ALGEBRA 


EXERCISE 59 
CUMULATIVE REVIEW 


1 . Factor: (a) x^°' — 13 + 36 ; (c) 12 — 27 

(b) Q xy — 35 2 /^; (d) x^^ — 2 x^y^ + y^ 

2. Write as a single fraction in its lowest terms : 

/i 4 - 5 g + 4 \ ^ /-. , a - 16 \ ^ / g^ + 6 g + 8 \ 

\ g^ — 3 g — 4/ \ a? — a) \ g^ + g / 


3. Find the numerical value of 


6 g^ — g 


3 g “1“ 2 


Solve the formula - = ^ 


for L. Find the value of L, 

g h -\- L 

correct to the nearest tenth, when a = 5.2, g = 32.16, and 
h = 7.5. 


6. Reduce to lowest terms : 


2 a — b , 2 g + ^ 
g + 36 a — 3b 


2 a — b _ 2 g + 6 
g + 3 6 


Check. 


(o) 


a — Zb 

Solve the systems of equations : 

—- H-= 0 fl2m—4w + n = 3 

^_1 1 

^ ' = 0 


m—w—2p=— 1 
5 m — 2 w = 0 


2a:-3 2y -^IZ 

7. Find two numbers whose sum is m and difference is n. 

8 . The sum of the three digits of a certain number is 13. If 
the number, decreased by 8, be divided by the sum of its units’ 
and its tens’ digits, the quotient is 25; if 99 be added to the 
number, the result has the same digits as the original number, 
but in reverse order. Find the number. 

9. Solve the systems : 


(a) 


X — ay = ab — 2 a 
bx ay = — ab 
10. Solve the formulas mg ■ 


(b) 


mx — ny = 0 
mn(x — y) — 

T = ?7i/and T = n/for T andf. 
















VII. SQUARE ROOT AND QUADRATIC SURDS 

110 . To study the next chapter intelligently, you must know 
how to find the square roots of certain numbers. Possibly 
you remember how to do this from your previous course in 
mathematics. The following diagnostic test will help you dis¬ 
cover what you do remember and what you must study with 
special care. 

DIAGNOSTIC TEST 

1. What is the square root of 25? Why? 

2. Is there more than one square root of 25? 

3. Can you give the positive square root of each of the 
following: 81? 144? 196? 169? 121? 225? 256? 

4. WhatisV64a2? 

5. What is (36 a:^)^? 

6. Find the square root of 2304. 

7. Find the square root of 137 to two decimal places. 

8. If v/O = 2.45, how much is Vl50 to two decimal places. 

9. If VlO = 3.162, how much is Vf, to two decimal places. 

10. If V6 = 2.45, how much is J to two decimal 

places. 

Note. If you cannot do all these examples you need some of the 
Remedial Instruction and Practice which appears on pages 100 to 104. 
If you can do all these examples you can spend more time on the new 
material which appears on pages 105 to 113. 

Note to the Teacher. Only square roots appear in this chapter. 
Chapter XII can be studied now if the teacher wishes to teach cube root 
at this time. 


99 



100 


ALGEBRA 


Remedial Instruction and Practice 

111. The square root of a number is one of two equal factors 
whose product is the given number. 

Since (+ 3 a)(+ 3 a) = + 9 + 3 a is a square root of 9 a^. 

Since (— 3 a)(— 3 a) = + 9 a^, — 3 a is a square root of 9 a^. 

Every number has two square roots. They are written to¬ 
gether by means of the double sign, zt, read plus or minus. 

The positive square root is called the principal square root. 
The square root means the principal square root. 

The principal square root of a number is indicated by the 
radical sign, V” , or by the exponent 

So V 25 = +5, and also 25^ = + 5. 

The number under the radical sign is called the radicand. 

112. Recognizing some perfect squares at sight. 

(a) Perfect square arithmetical numbers. You should memo¬ 
rize, if you have not, the squares of the numbers 1 to 15, or 
1 to 20. The more you know the better. 

{h) Perfect square monomials have perfect square arithmetical 
coefficients and literal factors which have even exponents. 

113. First principle employed in extracting roots. 

The square root (or any other root) of a product of two or more 
numbers is the product of the square roots of the numbers. 

Thus: V1746 = V4 • 441 = Vi • VmT • Vi^ = 2 • 21 • a, 

or 42 a. 

EXERCISE 60. REMEDIAL PRACTICE 

Give the principal square root of: 


1. 

49 

6. 100 xy 

9. 900 a^b^ 

13. 

196 m2 

2. 

64 a^ 

6. 81 

10. 225 m2 

14. 

1089 .t6 

3. 

121 

7. 169 a® 

11. 2500 

15. 

1936 a262 

4. 

144 7-2 

8. 400 

12. 625 a^ 

16. 

2304 m^ 




SQUARE ROOT AND QUADRATIC SURDS 101 


114. The square root of a perfect square polynomial can be 
found by a long division process. (Optional) 

Example. Find the square root of 9 + 25 — 30 xy. 


Solution. 1. Arrange the polynomial in descending powers of x. 

2. VQ = 3 X. Write 3 x in the root. Root 3 x — 5 y 

3. Square 3 x, getting 9 x^. Write 9 x^ — 30x2/ + 25 

it below 9 x^. 9 x^ 

4. Subtract, obtaining the first re- 


mainder. 

5. Trial divisor, 2 • 3 x = 6 x. 6 x 

— 30 X2/ -r- 6 X = — 5 y. 

Write — 5 ^ in the root and add 

— 5 2/ to 6 X, forming the complete — 5 2 / 

divisor. 6 x — 5 2 / 

6. Multiply the complete divisor by 

- 5 y. 


— 30 X2/ + 25 2/2 


— 30 x^ + 25 y^ 


7. Subtract, In this case, no remainder. 

8. The square roots are + (3 x — 5 y) and — (3 x — 5 2 /). 


EXERCISE 61. (Optional) 


Find the square root of: 

1. 25 + 40 xy + 16 y- 

2 . 9 - 30 ab + 25 ¥ 

3. 36 c2 - 60 cd + 25 d? 

4. 49 — 28 m?n 4 

5. 25 — 70 rs + 49 


6. 4 -h 4 + 5 + 2 X + 1 

7. 4 x^ — 4 x^ + 5 x^ — 2 X -1- 1 

8. 9 x"* + 6 x^ -|- 13 x^ + 4 X -h 4 

9. 9 + 12 _ 2 - 4 a -h 1 

10. — 6 + 13 2 /^ — 12 2 / + 4 


11. 4 - 12 cH + 17 - 12 cd^ + 4 d^ 


12. 9 — 24 mhi +10 + 8 mn^ + 

13. 1 - 2 X - x^ + 2 x^ + x4 

14. a“ + + + + 2 a6 + 2 ac + 2 6c 

15. 4 + 9 62 + c2 + 12 a6 - 4 ac - 6 he 

16. 9 a2 - 24 ah + 30 ac + 16 h" - 40 6c + 25 c^ 

17. 4 x2 - 12 xy - 16 xz + 9 if + 24 2 /z + 16 z^ 
Note. Not required by C.E.E.B. 








102 


ALGEBRA 


115. Finding the square root of an arithmetical number. 

Example 1. Find the principal square root of 4624. 

Solution. 1. Since 4624 is more than 3600, V4624 is more than 60; 
and since 4624 is less than 4900, 4624 is less than 70. 





a 


h 



60 -b 8 


6 8 




46 24 


46 24 

1. 

QQ 2 = 3600. Subtract from 4624. 


36 00 


36 

2. 

2 X 60 = 120; 1024 ^ 120 = 8+ 

120 

10 24 

120 

10 24 

3. 

Add 8 to 120 and to 60. 

8 


8 


4. 

Multiply 128 by 8. Subtract. 

128 

10 24 

128 

10 24 


Note 1. Use only the form h when writing the solution. 

Note 2. 4624 was separated into groups of two figures each because 
the square of any number contains as many groups of two figures each 
as there are digits in the given number — except that the left-most 
group may have only one digit in it. 

Thus: 32 = 9; 8^ = 64; 12^ = 144; 95^ = 9025. 

Similarly 34,038 will be written 3 40 38. This shows that its square 
root has three figures to the left of the decimal point. 

A number written in decimal form is divided in the same manner, 
counting in both directions from the decimal point. Thus, 34256.895 
becomes 3 42 56.89 50. The square root of this number has three 
figures to the left of the decimal point. 


Example 2. Find the principal square root of 5207.0656. 


Solution. 1. The largest square number in 52 is 
49. Write V'49 or 7 in the root. Subtract. An¬ 
nex 07. 

2. Trial divisor is 2 X 7. Annex 0. 307 -J- 140 
= 2+. Place 2 in the root. Add 2 to 140. Multiply 
142 by 2. Subtract. Annex 06. 

3. Trial divisor is 2 X 72. Annex 0. 2306 
1440 = 1+. Place 1 in the root. Add 1 to 1440. 
Multiply 1441 by 1. Subtract. Annex 56. 

4. Trial divisor is 2 X 721. Annex 0. 86556 
14420 = O'*'. Place 6 in the root. Add 6 to 14420. 
Multiply 14426 by 6. No remainder. 

Note. The root, correct to tenths, is 72.2. 


7 2.1 6 
52 07.06 56 
49 


140 
2 
142 

1440 
_ 1 

1441 
14420 

6 

14426 


3 07 


2 84 


23 06 


14 41 


8 65 56 


8 65 56 





















SQUARE ROOT AND QUADRATIC SURDS 103 


Rule. To find the principal square root of a number: 

1. Separate the number into groups {periods) of two figures 
each, startmg at the decimal point, and forming the groups each 
way from the decimal point. 

2. Find the largest square number not more than the left-most 
period. Write its square root as the first figure of the square root. 
Subtract the square number itself from the first period. 

3. To the remainder, annex (bring down) the next period. 

4. Form the trial divisor by doubling the part of the root already 
found, and annexing a zero. 

5. Divide the remainder formed in Step 3 by the trial divisor. 
Annex the quotient to the part of the root already found, and add 
it to the trial divisor to form the complete divisor. 

6. Multiply the complete divisor by the new figure of the root. 
Subtract the result from the remainder formed in Step 3. 

7. If other periods remain, repeat Steps 4, 5, and 6 until there 
is not a remainder, or until the desired number of decimal places 
for the root have been obtained. 

Note. In Step 6, if the product is greater than the remainder, the 
last figure obtained for the root is too large. 

116. Only the approximate square root can be found in most 
cases. For example, \/20 does not have an exact square root. 
In such cases, we carry out the square root to as many decimal 
places as we please, annexing pairs of zeros as they are needed. 

EXERCISE 62. REMEDIAL PRACTICE 

Find the principal square root, correct to hundredths: 


1. 

4489 

6. 

16,129 

11. 13 

16. 

4.7524 

2. 

8836 

7. 

27,556 

12. 91 

17. 

806.56 

3. 

5625 

8. 

11,664 

13. 165 

18. 

46.3761 

4. 

9604 

9. 

95,481 

14. 23 

19. 

723.61 

6 . 

6889 

10. 

214,369 

16. 463 

20. 

568.25 


104 


ALGEBRA 


117. A table of square roots can be used to advantage. A 
short table appears on page 253. Often the square root of a 
number not in the table can be found by using a number in 
the table. 

Thus: Vi^ = V4 • 33 = Vi • V^ = 2 X 5.7446, or 11.4892- 
• Correct to hundredths, Vl32 = 11.49. 

You will find it a big help if you memorize the three place 
values of V2, Vs, Vfi, and Vfi. (See page 253.) 

118. The square root of a fraction equals the square root of 
its numerator divided hy the square root of its denominator. How-- 
ever, always change the fraction to an equal fraction whose 
denominator is the smallest perfect square into which the given 
denominator can be changed easily. 

Example. ^1 = ^5 = ^ = = 1.224 = 1.22 


Find the following correct to hundredths: 


1. 

V 20 

3 . Vl08 

6. Vl25 

7. Vl75 

9. V 18 O 

2. 

V 27 

4. Vn2 

6. ViX) 

8. Vi47 

10. Vm 

11. 


13. Vf 

15. V| 

17. V* 

19. V* 

12. 


14. V| 

16. V* 

18. 

20. 


21. Find the simplest radical form and also the decimal value 
correct to hundredths of 


0 7. 1 

/l 1 /2 1 V 2 

2 - 

- 3 V 2 


Solution, g — 

' H 

1 

ICO 

II 

1 

ICO 

II 

l(N 


6 


Since V 2 = 1.414, 

2 - 3 V 2 2 - 4.242 - 

6 6 “ 

2.242 

6 

= - .373 = - .37. 

22 . i - Vf 

26. -i-fVi 

30. 

_ 1 

5 

+ Vi 

23. i + 

27. - I - Vi 

31. 

-4- 3 
' 8 

-V 

24. 1 - Vi 

28. f - V| 

32. 

_ 5 

6 

+ v^ 

26. f + Vi 

29. -i+Vf 

33. 

+ A 

- V* 








SQUARE ROOT AND QUADRATIC SURDS 105 


Quadratic Surds 

119. A quadratic surd is the indicated square root of a 
number which is not a perfect square; as Vs. 


120. A surd expression is an expression which involves one or 
more surds. In this chapter, only simple quadratic surd ex¬ 
pressions will be considered. 

121. A surd quadratic is in its simplest form when the num¬ 
ber under the radical sign is not a fraction and does not have 
any perfect square factors. 


122. To simplify a surd expression, first simplify the surds 
and then perform the indicated operations. 

Example 1. Simplify V 20 + V45. 


Solution. 1. + V 45 = V 4 • 5 + 

2 . = 2 V 5 + 3 V 5 , or 5 V 5 . 

This is the resultjn simplest radical form. 

3. .-. V 20 + V 45 = 5 V 5 = 5 • 2.236, or 11.180. 

Note. The surds 2 V 5 and 3^/5 are called similar surds, because 
their surd factors are the same. Only similar surds can he combined. 


Example 2. 


Simplify the surd 


s/ 


(?d 

a -\- h 


Solution. 


{ 


cH 

a + & 



or —f^Vcd(a + h). 

0 + 5 


Example 3. Find the simplest radicaj_ form and also the 
decimal value correct to hundredths of Vf + V-l-. 


Solution. 1. 




16 
3V2 


V2 












106 


ALGEBRA 


EXERCISE 64 

Find the simplest radical form, and also the decimal value of 
the following expressions, correct to hundredths : 


1 . VI 2 +V 75 

9. 


17. 

-i+A 

2. —VI 8 

10. 


18. 


3. VSO -V^ 

11. 

+ i -'A 

19. 

- 1 -As 

4. Vi5 +'y% 

12. 

+ i -A 

20. 

+ f - Vi^ 

1 

1 

id 

13. 

-i+A 

21. 

A+'^^ 

6. 3 V 27 -Vis 

14. 


22. 

a/s _a/ 1 

7. 2V^ +Vra 

16. 

+ i+Vf 

23. 

Vf -V24: 

8. 2 V^ -V 45 

16. 

_ 3 . —Vx 

24. 

V^ — a/a 

26 . 2 V^ -V 45 +V 5 

27. VI 

-A 


26. -V^ +V 72 28. Va d-V'aV 


Find the simplest radical form: 


29. 

V^ +V 16 a — V 25 a 

36. 

+ 4 c 

- V 16 e‘ - 

-4d2 

30. 

V 9 x^yz 

— 

—Vx^yz 

V 9 ax^ + V 4 


4 


37. 

ab —Vab^ — ¥ 


31. 

2b 


32. 

2V^ 

-VW^ + 5V^ 

38. 

— 6c- 

-Vmc^ -h nc^ 

33. 

V 180 - 

-V^ + 3VWa 



2c 


39. 

at - 

■Vaf — t^ 


34. 

CO 

00 

1 

■VYKa^+S aV^ 


at 


35. 

- 2 +V8 a2 - 4 

40. 

-2b - 

-V 4 b‘‘ - 

4 ac 


2 


2 a 


41. 

V 

\^,2 

44. 

^'2^ 

47. 

V“ 

^TT 

60. 

tf 

42. 


46. V 

48. 


61. ^ 

1 

^x - y 

43. 

V? 

46. 

^2 r 

49. 


62. 

la-b 
la b 




















SQUARE ROOT AND QUADRATIC SURDS 107 


123. Multiplication of simple quadratic surds. 

Since Vab = Va • Vb, then also Va • Vb = V^. 

Thus, V2 • Vs = V2^, or Vq. 

Example. Find the simplest radical form and also the 
decimal value of 2 V 3 X 5V2 correct to hundredths. 

Solution. 1. 2 V 3 X 5 V 2 = loVs^ = lOVe. 

2. IOV 6 = 10 X 2.449 = 24.49. 

Note. Keep the product under the radical sign in factored form. 
Thus, VI 8 • V 12 = VI 8 • 12 = V9 • 2 • 4 • 3 = 3 • 2 Vo, or oVo. 


EXERCISE 65 


Find the simplest radical form and also the decimal value of 
the following, to hundredths: 

1 . \/2 • Vs 4. Vo • VTo 7. 3V2 • 4V5 10 . ( 2 V 3)2 

2 . V3 • Ve 6 . V7 • Vli 8. 2V6 • 5V2 11 . ( 3 V 2)2 

3. V5 • Vio 6. Vn • V^ 9 . 5 V^ . 3 VIo 12 . (iV 3)2 


13. Multiply 3 + 5 V 2 by 3 + 5 V 2 . 

Solution. 3 + 5V^ 

3 + 5 V 2 

9 + 15 V 2 

15 V 2 + 25 V 4 


9 + 30 V 2 + 25 • 

Find: 

14. (5 - Vs) (5 + Vs) 

16. (5 + V6)(5 - Ve) 

16. (3 - V2)(3 + V 2 ) 

17. (9 - 2V5)(9 + 2 V 5 ) 

18. (- 7+ VIDC- 7 + Vl3) 

19. (10 - 3V6)(10 + 3 V 6 ) 

20. (3 - 2-^/2) (3 + 2\/2) 

21 . (5 + 3V7)(5 - 3 V 7 ) 


2 or 59 + 30 V 2 

22. (Vx - 5)^ 

23. (V 2 X + 3)* 

24. (V- 5 + 

26. (2 - Vsy 

26. (a — Vly 

27. (m + Vr)^ 

28. (V V — A acY 

29. (o — ’'/be) (a + Vfcc) 










108 


ALGEBRA 


124. Division of simple quadratic surds. 

Rule. The quotient of the square roots of two numbers 

equals the square root of the quotient of the numbers. 

Example 1. 6V^28 2'V^7 = = 3V^4 = 3 • 2, or 6. 

Example 2. VE H- VlE = = Vf = V|, or jVs. 


EXERCISE 66 

Find the following quotients. 


1. Vli -H V7 

2 . VE6 ^ Vio 

3. V72 V2 

4 . V45 Vs 
6 . VI^ ^ Vs 

6 . sV^ + Vs 

7. iVn 2 V 3 


8 . eVei vT? 

9 . 2V40 Vi 

10 . sVm Vs 

11. V^ 2V3 

12 . V32 3V2 

13. Ve VS 

14. V 7 V 35 


16 . Vis H- 3 V 5 

16. 2 V 99 + sVIT 

17 . sVi - Ve 

18. 4 V 3 Vs 

19. 3 V 2 H- Vi 2 

20 . 11 Vs Vis 

21. sVe 4 Vi 8 


125. Making a monomial divisor a rational number. 

^ , 20 20 • Vs 20V3 20V3 10V3 

Example. ^ = Vfi = ^- 

The numerator and denominator are both multiplied by a 
surd which makes the new denominator the square root of the 
smallest possible perfect square. 




EXERCISE 

67 




2 

5 . 

15 

9. 

19 

13. 

6 

V 3 

V3 

Vl 9 

V^ 

4 

6. 

8 

10. 

999 

14. 

14 

V5 

V12 

Vg^ 

V^ 

6 

7. 

9 

11. 

7 

16. 

25 

V2 

V18 

Vii 

V^ 

12 

8. 

20 

12. 

2S 

16. 

33 

Ve 

VIo 

VI5 

Vn 








SQUARE ROOT AND QUADRATIC SURDS 109 


126. Rationalizing a simple binomial divisor. 


Example. 


Rationalize the denominator of 


3V2 +1 
2V2 - 1 


„ , , 3 V 2 + 1 ( 3 V 2 + 1 )( 2 V 2 + 1 ) 12 + 5 V 2 + 1 

Soluticm. 1 .-^- 7 =-- 7 ^-^-^ ---• 

2 V 2 - 1 ( 2 V 2 - 1 )( 2 V 2 + 1 ) 8-1 

^ . 3 V 2 + 1 13 + 5 V 2 13 + 5(1.414) ^ ^ 

• •• " 7 - 7— - 

Note. The surd expressions 2 V 2 + 1 and 2 V 2 — 1 are called 
conjugate surds. 


Rule. To rationalize a binomial surd denominator, multiply 
both numerator and denominator by the conjugate of the denom- 


inator. 

EXERCISE 68 



1 . 

6 

9. 

V2 +1 

17. 

Vs + 5 

3 + V5 

V2 - 1 

Vs - 5 

2 . 

5 

10 . 

2 - V3 

18. 

V7 - 2 

Vl - 2 

2 +Vz 

V7 + 2 

3. 

4 

11 . 

3 - V2 

19. 

2V2 - 1 

3 - V7 

3 + V2 

2V2 + 1 

4. 

6 

12 . 

4 + V5 

20 . 

2V3 - 1 

(M 

1 

ICO 

> 

4- VE 

V3 - 1 

5. 

7 

13. 

6 + V2 

21 . 

2 - 3V2 

2V2 -1 

V2 - 1 

3 - V2 

6 . 

11 

14. 

8 - Vz 

22 . 

2V5 - 1 

2V3 + 2 

2 - Vz 

V5 + 2 

7. 

9 

16. 

10 + VE 

23. 

Vo - 2 

5 - 2V3 

z+ VE 

2V6+ 3 

8. 

10 

16. 

Ve - 5 

24. 

3V6 - 5 

6 - ZV 2 

Ve - 2 

2V6 - 1 

































no 


ALGEBRA 


Radical Equations 

127. A radical equation is an equation in which the variable 
appears under a radical sign or with a fractional exponent. 

The radical sign or the fractional exponent always indicates 
the principal root in a radical equation. 

Example 1. Solve the equation x — 1 — — 5=0. 

Solution. 1. X — \ — 5 = 0._ 

2. .•. X — 1 = Va;2 — 5. 

3. Squaring both members, — 2 x 1 = x^ — 5. 

4. — 2 a; = — 6, or a; = 3. 

Check. Does 3 - 1 - Vq - 5 = 0? Does 2 - Vi = 0? Yes. 

Note. When a single radical occurs in an equation, transpose 
the radical to one side and all other terms to the other side. Then, 
if the radical is a square root, square both members of the equation; if it 
is a cube root, cube both members; etc. 


Example 2. 


Solve the equation x — 1 — Va:^ — 5 = 0. 


Solution. 1. a; — 1 + Va;^ — 5 = 0. 

2. .•. + Va;2 — 5 = 1 — a;. 

3. Squaring, — 5 = *1 — 2 a; + 

4. .'. 2 a; = 6, or a; = 3. 

Check. Does 3 - 1 + Vg - 5 = 0? Does 2 + Vi = 0? No. 

Therefore 3 is not a root of the given equation. 

What is the explanation of the solution a; = 3? If the original 
equation is compared with the equation of Example 1, it is noticed 
that the only difference is in the sign of the radical; also that in Step 3, 
after squaring both members in both examples, the resulting equations 
are the same. In each example, if the equation of Step 1 has a root, that 
number is a root of the equation of Step 3; but, since the equation of 
Step 3 is the same in each solution, it cannot be asserted in advance 
whether its root or roots are roots of the equation of Example 1 or of 
Example 2. When finally the solution a: = 3 is obtained, the question 
arises, is 3 a root of the equation in Example 1 or in Example 2? The 
root X = S satisfies the equation of Example 1; it does not satisfy the 
equation of Example 2. In Example 2, the extraneous root 3 is i ntrodu ced 
by the method of solution. In Example 2, Step 2 really is Vx^ — 5 = 
— {x — 1). When squaring, the minus sign in front of (a; — 1) is lost. 











SQUARE ROOT AND QUADRATIC SURDS 111 


EXERCISE 69 

Solve and check the following equations: 


1. 

V7a+2-4 = 0 

11. 

Vl3 - 

- 4x = 2x + 11 

2. 

1 

II 

CO 

1 

12. 

V5 

— 6 X + 9 = 10 

3. 

V7a:-34-3 = 2a: 

13. 

V9x2 

— 5x+l = 3x 

4. 

Vx - 1 = 2 

14. 

V3x+l = Vll + x 

6. 

Vx + 1 = - 3 

16. 

VIO - 

- 3 X = 4 — X 

6. 

V2x - 5 = + 3 

16. 

V4 X - 

f 5 = 3x + 4 

7. 

-Vx+3 = - 3 

17. 

Vi - 

5i/ + 2/ = 1 

8. 

— Vx + 4 = 4 

18. 

V8x+5 — l = 4x 

9. 

Vl - X = 2 

19. 

V15x 

+ 11 = 3(x + 1) 

10. 

CO 

1 

II 

(M 

1 

1 

> 

1 

20. 

V'l2x 

+ 7 = 12x - 5 



(a) for I (6) for g 

22. Solve the equation V = ^2 gs 
{a) for g (6) for s 

23. Solve for 5: ^ ^ 

24. Solve for A: L = irrV 


Solve for x : 

Suggestion. First clear of fractions. 




_ 9a2-62 


2 a 

25. 

Vx — 12 a6 = 26. 

Va + x — V2x = 

Va + X 

27. 

V X + 3 + 

28. 

Vx Vx 

Vx + Vx — 9 = 

36 

Vx — 9 

OQ 

10 X 

: + VlO X + 2 - 

1 

2 


AM. 

VlOx - 8 

VlO X — 9 



vr+1 _ 

V< + 3 



30. 

V< + 3 

Vi + 6 




Note. Optional examples appear on p. 245. 












































112 


ALGEBRA 


128. {Optional.) There is a functional relation between a 
number and its square root, because for every number there is 
a square root. In fact there are two square roots. This is some¬ 
thing new in a functional relation. It is interesting to study 
it graphically. 

Let y = x^. 

This means that x = zb 

When a; = 0d=2 =L4 =L6 d=8 ±10 

then y — 0 +4 + 16 + 36 + 64 + 100 


Y 



Notice: 1. For every value of x there is one value of y. 

Thus when a: = 7, y = 49. 

2. But for every value of y there are two values of x. 

Thus let y = 56. From point C, go to the right to point + 
on the graph, and then down to the a:-axis at the point whose 
abscissa is about + 7.5. This is the positive square root of 56. 

Similarly go from point C to the left to point D on the graph, 
and then down to point E whose abscissa is about — 7.5. This 
is the negative square root of 56. 

Note. For convenience the ?/-unit is one tenth as large as the x-unit. 
This flattens the curve considerably. 





































































































SQUARE ROOT AND QUADRATIC SURDS 113 


1. Solve for x\ 


EXERCISE 70. CUMULATIVE REVIEW 

x + 1 o: + 4 


2. Solve for x: 3 — 


a: + 3 
5 - 


X + 2 

2x , 4 

= 4- 


Check. 


+£± 2 , 


5 10 

3. Solve for x: Z cx — b a -\- h — 2 c = 66 — (a+3 6x + 2c). 

4. The sum' of the digits of a number of two digits is 9. 
If the digits be reversed, the new number is only three eighths 
as large as the old. Find the number. 

6. A crew can row 4 miles against the current in forty 
minutes and the same distance back again in twenty minutes. 
What is the rate of the current and of the crew in still water? 

6. One hundred pounds of an alloy of silver and copper 
contains 2 parts of silver to 3 of copper. How much copper must 
be melted with the alloy so that the resulting alloy will contain 
3 parts of silver to 5 of copper? 

6 X — 5 y = 25 
4 X — 3 y = 19 

What kind of equations are: 


7. Solve the system 

8 . 


f22/-3x = 0 
^“M3:^-22/ = 5 

9. Solve the system 


3 X 

6 X 


10 . 


(fc) 

ax hy = G 
px - qy = ^ 

W , 

(a) Solve the formula D = 


2 y = 5 
4y = 10 


(6) If TU is constant, and D increases, how does w change? 

11. Find, correct to hundredths ^ “ \/| 

px-2y + 4s = 13 

12. Solve the system: ^ 2x + 5y — 3z = —9 

[0x + Sy4-2z = 7 







VIII. QUADRATIC FUNCTIONS AND EQUATIONS 

One Variable 

129. An interesting formula from Physics. • You know the 
saying that anything that goes up must come down.” The 
formula S = at - IQ f will help you find approximately how 
high up ^‘it” will go, how long it will take to get there and when 
it will “come down,” provided you know the rate at which it 
started “up.” In this formula: 

a = the rate in feet per second at which the object is 
started on its upward journey; 
t = the number of seconds it travels; 

S = the distance it goes in t seconds. 

Let us take the case of a ball which is thrown upward at the 
rate of 100 feet per second. 

Then S = 100 t - IQ t\ 

Evidently the distance S depends on the time t. 

When t = 0, S = 0, since the ball is still on the ground. 


When t = 

0 1 

2 

3 

4 

5 

6 

7 

then S = 

0 84 

136 

156 

144 

100 

24 

- 84 


The graph on page 115 is the result of locating points whose 
coordinates are the corresponding pairs of numbers in this table. 

Remember that, on this graph, S represents the distance up 
from the ground, and that t is the number of seconds it takes 
the ball to travel this distance S. 

Thus, at the end of 1 second, the ball is up 84 feet. At some time 
between 3 and 4 seconds, the ball reaches its highest position, which is 
a little more than 156 feet. 


114 





QUADRATIC FUNCTIONS AND EQUATIONS 115 


Question 1. When does the ball reach the height of 100 feet? 
That is, for what value of < is 100 ^ — 16 = 100? 

Answer. At A, S = 100 ft. and t = about 1^ sec. 

At B, S = 100, and t = 5. 

So 1^ and 5 both satisfy the equation 100 t — IQ f = 100. 
Question 2. When is the ball up 24 feet? 

That is, for what value of i is lOO f — 16 = 24? 

Answer. AtC,S = 24 and t = at Z), ^ = 6. 

Both of these values satisfy the equation 100 ^ — 16 = 24. 

Question 3. When is the ball on the ground? 

That is for what value of Hs 100 i — 16 = 0? 

Answer. At E, S = 0, and t = about 6:|-. 

Also when < = 0, S = 0, or 100 ^ — 16 = 0. 

There are two values of t which satisfy each of the equations. 


Sft. 



















































































































116 


ALGEBRA 


130. loot — 16 is a second degree function of t. 

The equations 100 t — 10 f = 100; 100 t — 16 = 24; and 

100 t — 10 = 0 are second degree or quadratic equations. 

Observe that the variable t does not appear in a denominator 
or under a radical sign and that its highest exponent is 2. 

Since there is a term containing the first power of t, 
100 f — 10 t‘^ = 100 is a complete quadratic equation. 

Similarly + x — 5 = 0isa complete quadratic equation. 
But 3 — 5 = 0 is an incomplete quadratic equation. 


131. Solving the incomplete quadratic equation — 9 = 0. 

(a) Graphical solution. 1. Let y = — 9 


When X = 

-4 

-2 

-1 

0 

1 

+2 

+4 

Then y = 

+7 

-5 

-8 

-9 

-8 

-5 

+ 

2. At A: 

y = 

0; : 

c = 

- 3 . 




Does ( — 

3)2- 

■ 9 = 

= 0? 

Yes 




At jB: ?/ 

= 0; 

X = 


h 





Does 32 — 9 = 0? Yes. 

The equation — 9 = 0 has the two 
roots + 3 and — 3. 

(6) Algebraic Solution. 1. — 9 = 0 

2. Ag: x^ — 9 


Y 



3. Take the square root of both sides. Then ± a; = ±3. 

This looks like four equations (1) + a; = + 3; (3) — a; = + 3. 

(2) + a; = - 3; (4) - a: = - 3. 

4. M_i(3) Then + a: = — 3. This is equation (2). 

M_i(4) Then + a; = +3. This is equation (1), 


So we actually get only two equations, and we get these if we use the 
+ sign on the left in Step 3, and dr on the right. 


Rule. To solve an incomplete quadratic equation: 

1. Simplify the equation until it takes the form = a number. 

2. Take the square root of both members of the equation, placing 
the + sign in front of the square root of the left side and the double 
sign (db) before the principal square root of the right side. 


























































QUADRATIC FUNCTIONS AND EQUATIONS 117 


EXERCISE 71 

Find the roots, correct to tenths: 

1. 11 - 176 = 0 4. 10 - 2 = 88 - 5 

2 . 5 - 120 - 22 6. 7(t - 3) t{t - 5) =2t 

^ 9 + x 
Sx-\-2 


3. +2 

+ *-2-3 6.2^ 

X - 2 

a: + 2 4 ■ 

7. Solve for a;: a — 2 cs? = 8 b, 

Solution. 

1. a — 2cx^ = Sb. 

2. Sa 

- 2 cx2 = 3 & - a. 

3. D_2c 

^2 _ a - 3 6 
* 2c • 

-H 

II 



a. Solve each of the following equations for the literal number 
having exponent 2. 

b. Find the value of this number correct to tenths for the 
given values of the other numbers. 

8. (a) A x^ = m. (6) Find a:, if ^ = 6 and m = 90. 

9. (a) cx^ = d. (6) Find a:, if c = 12 and d = 8. 

10. (a) S = ^ gf. (6) Find t, ii S = 1608 and g = 32.16. 

11. (a) f = (b) Find v, if/ = 3750, r = 5, and m = 125. 

r 

12. (a) V = ^Trr^H. (b) Find r, if U = 840, tt = and 
H = 21 . 

13. (a) S = 4 7rr^. (b) Find r, if 8 = 576 andTr = 3.14. 


14. 3 + l 

16 82/ -1 

2/ + 1 

2 3 

4 2/ + 1 

32/ + 1 

. x^-\-x-\-\ a;2—a: + l _ 0 

17- ^ 0 + 

U=7 


t 2 

t - 1 




















118 


ALGEBRA 


132. Solving a complete quadratic equation graphically. 

X \ 2 

Example. Solve the equation o ~ o — “' 

o S X 


Solution. 1. Mss. Transpose, — x — ^ — 0. 


2. Let y = x^ — X — 6 


When X 
is 

Then y 
is 

- 4 

+ 14 

- 3 

+ 6 

- 1 

- 4 

0 

- 6 

+ 2 

- 4 

+ 4 

+ 6 

+ 5 

+ 14 



3. The graph is above. It is called a parabola. 

4. y is zero for all points on the a:-axis. Therefore the function 
x^ — X — Q = 0 at all points where the parabola crosses the a:-axis. 

At A: X = - 2. Does (- 2)2 _ (_ 2) - 6 = 0? Yes. 

At R: X = + 3. Does (3)2 - (3) - 6 = 0? Yes. 

— 2 and + 3 are the roots of a:2 — a; — 6 = 0 

Rule. To solve an equation having one variable graphically: 

1. Clear the equation of fractions; collect like terms; and 
transpose all terms except zero to the left side of the equation. 

2. Represent by y the function obtained in Step 1. 

3. Draw the graph of the function in Step 1, making the vertical 
axis the y-axis or the function-axis. 

4. At each point where the graph crosses the horizontal axis, 
y is zero, and the abscissa of the point is a root of the given 
equation. 

Note. The rule furnishes a graphical solution of an equation having 
one variable not only when the equation is of the second degree, but for 
an equation of any degree. 









































































QUADRATIC FUNCTIONS AND EQUATIONS 119 


EXERCISE 72 


Solve the following equations graphically. 


1 . X - 12 = 0 

2. x^ — 6 X = IQ 

3. x‘^ — X = 12 

4. = 10 + 3 a; 


6. a:^ + a: = 20 

6. 2 a:2 + 3 a; = 20 

7. 2 a:2 = 7 a: + 15 

8. 4 a;2 - 25 = 0 


Find to the nearest tenth the roots of: 

9. a:2 + 2 a: - 5 = 0 12. a:^ + 3 a; - 11 = 0 

10. a:2 - 3 a: - 2 = 0 13. 2 a:^ - 3 a; - 3 = 0 

11. a;2 - 4 a; = 2 14. 2 a;^ + x = 11 

16. (a) Draw the graph of the function x^ — 3 x. 

(6) From this graph determine the roots of the equation 
a;^ — 3 a; = 0. 

(c) Determine also, from this graph, the roots of the equation 
a;^ — 3 a; = 4. 

Method. Start at the point ?/ = 4, on the ?/-axis. Imagine a line 
through this point, parallel to the x-axis. It cuts the graph at two 
points. The abscissas of these points are the roots of the equation 
— 3x = 4:. Determine them, and check them by substituting them 
in the equation. 

(d) Determine from this same graph the real or approximate 
roots of: 

(1) a;2 - 3 a; = 10; (2) a:2 - 3 a: = 18; (3) a;^ - 3 a; = 7. 

16. (a) Draw the graph of the function a;^ + 2 a:. 

(6) From this graph, determine the real or approximate roots 
of the following equations : 

(l)a:2+2a: = 0 (2)a:2 + 2a: = 3 (3)a:2 + 2a: = 8 

(4) a;2+2a:-15 = 0 (5) a^2_j_2a: = 10 (6) a;2-|-2a:-6 = 0 

17. (a) Draw the graph of the function x^ — Qx. 

(b) From this graph, determine the real or approximate roots 
of: 

(l)a:2«0a;4-5 = O (2)x2_0^^9 = O(3)a:2_0a. = lO 


120 


ALGEBRA 


133. Solution of complete quadratics by factoring. 

Example. Recall the formula S = 100 t — 16 on page 114. 
How long does it take the ball to rise 100 ft. ? That is, find t so 
that 100 = 100 ^ - 16 t\ 

Solution. 1 . Transpose 16f^ — 100i + 100 = 0 

2. D 4 4^2 - 25 ^ + 25 = 0 

3. Factor. ( 4 ^ _ 5 )(^ _ 5 ) =0 

4. If now 4 i — 5 = 0, then (4 ^ — 5) (i — 5) also = 0. 

But 4 ^ — 5 = 0, if 4 i = 5, or f = ^ • 

5. Similarly if f — 5 = 0, then (4 i — 5) (i — 5) = 0 . 

But ^ — 5 = 0, if ^ = 5. 

These are the roots of the equation as we know from page 115. 

134. A fundamental principle used. 

If a product is zero, then one or more of its factors is zero. 

Thus, if (x — l)(a; + 2) = 0, rc — 1 = 0, or a: + 2 = 0, or both 

a; — 1 = 0 and x + 2 = 0. Obviously, if neither is zero, their product 

is not zero. 

Example 2. Solve the equation — 2 M = 35. 

Solution. 1. Transpose — 2 M — S5 = 0. 

2. Factor. {M - 7)(M + 5) = 0. 

3 . AT = + 7, and M = — 5. 

Note. In Step 3, part of the work was done mentally. 

Namely: (M - 7) = 0, if M = 7; (M + 5) = 0, if M = - 5. 

Check. These roots check when substituted in the equation. 

Also, notice: (a) that (+ 7) + (— 5) = — (—2) for + 7 — 5 = 
+ 2 , and that — 2 is the coefficient of M. 

That is, the product of the roots equals minus the coefficient of the 
first power of the unknown. 

( 6 ) that (+ 7)(— 5) = — 35, the third term of the equation. 

That is, the product of the roots equals the term free from the 
unknown. 

This check can be used only when the coefficient of the squared 
term is 1. Here is 1 • M^. If this coefficient is not 1 , divide both 
sides of the equation by it before using this check. 


QUADRATIC FUNCTIONS AND EQUATIONS 121 


Example 3. Solve the equation 2 - 3 ao; = 35 a?. 

Solution. 1. Transpose: 2 — 3 aa: — 35 = 0 

2. Factor. (2 a; + 7 a){x — 5 a) = 0 

3. a; = — f a; a; = 5 o 

Check. If the equation of Step 1 is divided by 2, then 

2 3a 35 , ^ 

^ =0. 

Now, does (5 a) = _ 

Does -'^a + ~a Yes. 

Also does (~ I ~ ^ 


EXERCISE 73 

Solve the following equations by factoring: 


1. 

j/2 - 12 2 / + 32 = 

0 

5. 

X2 - 

■ 11 X = 0 


2. 

+ 6 2 = 55 


6. 

8x2 

+ 5x - 3 

= 0 

3. 

= 63 + 2 m 


7, 

4 c2 

- 8 c = 21 


4. 

16*2 = 1 


8. 

3 x2 

II 

1 


9. 

i - X 12 


11. 

7 

, 1 _ 

3 

1 

CO 

1 

r-H 


3 - 

X 2 4 

— X 

10. 

7 1 y- 

2 

12. 

2 

6 

= 1 

1 

CO 

toi 

1 

1 

4 

3 - 

X 8 — X 


Solve the following equations for x: 


13. x"^ — 2 ax — = 0 

14. 4 a:2 - 9 m2 = 0 
16. 25 x^ = 16 a2 

16. 2 x^ — 7 ax 3 a^ = 0 

S a 5 2 a 

21 .-= - 

a:+6a 4 x — 5 a 

x + 2c ^ 36 

a: 3 c (a; 4" 3 c)2 


17. J x2 + ax — f a2 = 0 

18. ^x"^ = % — px 

19. i x2 + f rx = -y- r2 

20. f x2 + |- X5 = f 52 

^ 5m — 2x 6m 

23. - = - 

3m — 2x 3m — X 


24. 


16 t 

X — Q t 


X — 4:t 
X — 8t 


+1 


















122 


ALGEBRA 


Solution of Quadratics by Completing the Square 

135. The need for another method is proved by the attempt 

to solve the equation 100 ^ — 16 = 80. 

If you consult the graph on page 115, you find that t must be 
about 1 and about 5^. Neither of these satisfies the equation, 
although they almost satisfy it. So the graphical solution is 
unsatisfactory. 

Simplifying the equation, you get 4 — 25 i + 20 = 0. 

4 — 25 / + 20 cannot be factored, so the factoring method 

also is not satisfactory. We must therefore learn a new method. 

136. A preliminary review of perfect square trinomials. 

Preparation. 1. Write carefully in a column the following 
expressions and after each its expanded value. 

Thus: {x + 7)2 = + 14 a: + 49. Similarly for: 

(а) (a: + 4)^; (6) {y - 3)^; (c) {w + 6)2; {d) (z - 7)^; 

W (/) (w - i)2; (^) (r - 1)2; (h) (x - 1)2. 

2. The result in each case is a perfect square trinomial. 

3. Write the following expressions in a column; annex to 
each the necessary third term to make the resulting trinomial a 
perfect square. Then after it write the square root of it. 

Thus: for a:^ + 8 a;, write “a:^ + 8 a: + 16; x + 4.” 

(a) + Sx] (b) x^ - 12 a:; (c) a:2 - 20 a:; (d) x‘^ - 16 a:. 

In each case, did you take J the coefficient of x and square it? 

4. Do similarly for : (a) a:2 + f (How much is i of f ?) 

(б) w2 2 (^) y^ - ^y; (d) ^z; (e) t^ - ^t. 

5. Do similarly for: (a) a:2 + |-a:. (How much is | of f?) 

(6) (c) a:2 — i a:; {d) r2 + f r; (e) s"^ — § s. 

In parts 3 to 5 you have been completing the square for bi¬ 
nomials having a squared term with coefficient 1. To do so, you 
add the square of one half the coefficient of the first degree term. 


QUADRATIC FUNCTIONS AND EQUATIONS 123 

137. Solving a quadratic by completing the square. 

Example. Solve — 12 x + 20 = 0 . 

Solution. 1. S 20 : — I2x = — 20. 

2 . Complete the square of the left member by adding 36. 

A 36 — 12 a: + 36 = — 20 + 36 

3. .-. {x - 6)2 = 16 

4. Take the square root: x — 6 = it 4 

a: — 6 = + 4, or a; = + 10. This is one root. 

And a: — 6 = — 4, or a; =+ 2. The other root. 

Check. + 10, and + 2 satisfy the given equation. 

Rule. To solve a quadratic by completing the square. 

1 . Simplify the equation and transpose terms until the equation 
takes the form ax^ -\- bx = c. 

2 . If a is not 1, divide both sides of the equation by it, so that 
the equation takes the form x‘^ px = q. 

3. Now take one half of the coefficient of x; square it; add the 
square to both sides of the equation of Step 2. 

Note. In Step 3: to get •§• of a fraction, divide the numerator by 2 or 
multiply the denominator by 2. 

Thus: -I- of is -fj ^ of t 1® i^* 

4. Write the left side as the square of a binomial; express the 
other side in simplest {fractional) form. 

5. Take the square root of both sides, writing the double sign, 
zk, before the square root of the right side. 

6 . Let the left square root equal the + root of the right side 
obtained in Step 5. Solve the resulting equation. 

7. Let the left square root equal the — root of the right side 
obtained in Step 5. Solve the resulting equation. 

Check by substituting the results of Steps 6 and 7 in the given 
equation, if they are integers or fractions. 

If the results are expressed approximately as decimals, they 
will not satisfy the given equation. Then (a) the sum of the roots 
is minus the coefficient of x in the equation of Step 2 . 

( 6 ) the product of the roots is the third term in this same equation. 


124 


ALGEBRA 


EXERCISE 74 


Solve by completing the square. Check. 


1 . + 8 a: + 15 = 0 

2 . ?/2 - 8 2 / + 12 = 0 

3. 22 - 2 3 - 35 = 0 

4. a:2 + 2a;-3 = 0 
9. ic2 — 5a:+6 = 0 

(Remember | of 5 = f; and (f )2 = ^.) 


5. 2/2-42/4-5 = 0 

6. s2 = 6 2 4- 7 

7. w2 - 10 w 4- 9 = 0 

8 . ^2 4 - 85-9 = 0 


10. a:2 - 3 a; 4- 2 = 0 

11. 2/^ 4- 3 2/ - 4 = 0 

12. ^2 _ 5 2 = 14 

13. m2 4- m — 2 = 0 
18. Illustrative example. 


14. /2 4- 5 / = 6 
16. w2 - w = 12 

16. c2 - 7 c 4- 10 = 0 

17. a;2 4- 9 a: = 10 
a;2 4- 4 a: — 4 = 0. 


Solution. 1, a;2 4- 4 a: = 4 

2 . Complete the square: a:2 4-4a; + 4 = 4 + 4 

3. .-.(re+ 2)2 = 8 

4. Take the square root: a: + 2 = ± V8 

5. .-. a:i + 2 = + V8 a:2 + 2 = - VS 

Xi = — 2 + Vs a:2 = — 2 — VS 

Check. Does (- 2 +V8) + (- 2 - V8) = - 4? Yes. 

Does (— 2 + V8)(— 2 — V8) = — 4? Does + 4 — 8 = 4? Yes. 
Note. For this check, see page 123, end of Rule. 

To get the roots correct to tenths, use V8 = 2.828. 



xi = - 2 + 2.828; . 

•. aji = .828, or about .8. 



a:2 = - 2 - 2.828; . 

a:2 = — 

4.828, or about 

- 4.8. 

19. 

.V + 2 .T — 4 = 0 

26. 

tH 

1 

CO 

4- 

= 0 

20. 

to 

1 

to 

1 

1 —‘ 

II 

o 

26. 

2/2 - 3 1/ - 2 

= 0 

21. 

2 ^+ 42-3 = 0 

27. 

^2 + = 5 


22. 

— 4 w = 6 

28. 

22 - 2 = 3 


23. 

+ 6« - 3 = 0 

29. 

To 

1 

Ox 

1 

= 0 

24. 

— (>x — 1 

30. 

a:2 + 5 a: — 3 

= 0 



QUADRATIC FUNCTIONS AND EQUATIONS 125 


EXERCISE 75. (Optional) 
Illustrative Example. Solve ^ x = f 
Solution. 1. -\- jx = ^ 


2 . 

3- A(^) 

4. 

5. 

6 . 


I of I = i; (i)^ = ^ 

a:' + f a; + ^ = f + 

••• (a: + = H 


.-. X + 

i = ± iVii 

= - i + 

X2 = — i 

-1 + vn 

- 1 

- 5 

X 2 — 

- 1 + 3.3166 

- 1 

..Xi- g 

X 2 — 

2.3166 

- 4.3166. 

Cl = —=—; or xi = .46+ 

X 2 c ) 


1 - 3.3166 


;oT X 2 = — . 86 + 


Solve correct to hundredths. 

1 . x 2 + fx-7 = 0 

2 . — -|-a: + ^ = 0 

3. a:^ — fic — f=0 

4. a:^ + fa; — 2 = 0 
6 . y^-iy-i = o 
6 . 2 _ 1 = 0 


7. a;2+ia;-i = 0 
(Hint. § of ^ = 4 .) 

8. a:^ — f a; = 1 

9- 2/^ + f2/-3 = 0 

10. a:^ — f a: + ^ = 0 

11 . 22 + ^ 2 + i = 0 


12 . Solve the equation Sx^ — 5a: + 2 = 0. 


Solution. 1. 
2. D 3 


3x2-5x + 2 = 0 . 

- -I X + I = 0. 
(Complete the solution.) 


13. 2 x2 - 5 X + 2 = 0 

14. 3x2+ I0x + 3 = 0 

16. 6 x2 + X - 2 = 0 

16. 4 p2 — 2 2 ? = 1 

17. 5 r2 + 2 r = 7 

18. 3 m2 + 5 m — 2 = 0 


19. 3 x2 - 2 X - 3 = 0 

20. 8 2/2 + 4 2 / — 1 = 0 

21. 4 22>2 — 3 — 2 = 0 

22. 6 /2 + 8 ^ - 5 = 0 

23. 2 52 - 10 ^ - 9 = 0 

24. 10 22j2 + 4 222 - 3 = 0 









126 


ALGEBRA 


Solution of Quadratic Equation by a Formula 

138. All quadratic equations having one unknown can be 
put in the form ax^ + 6a: + c = 0. 

If we solve ax^ + hx c = 0 for x, the roots may be used 
as formulas for the roots of any quadratic equation. 


Solution. 1. 

2. Da 

3. Sc 

a 

4. 

5. A/ 62 \ 

V4 02 / 

6 . 

7. 

8. 

9. 


ax^ -\- hx c = 0. 

+ - • a; + - = 0. 

a a 

x^ + -• X = — 

a a 

1 rl^\ A. /AY = — 

2 [aj 2 a’ \2a) 4 

b , ¥ 

— ' X — 

a 4 G 


2,6 ,62 ¥ c 

X^ -\ - - X -T -j—5 ; — - • 

a 4 a^ 4 a 

¥ — 4 ac 




4a2 

'^¥ — 4 ac. 




2 a 

^¥ - 4ac 


2 a 


-h± - 4 


2a 


Example. Solve 2 — 3 x — 5 = 0, by the formula. 

Solution. 1. Comparing this equation with ax^ + 6x + c = 0, 
a = 2, b = — 3, c = — 5. 


2. The formula is x 

3. Substituting, x = 

4. .-. X = 


— 6 ± \/¥ — 4 ac 
2 a 


- (- 3) ± V(- 3)2 - 4r2)f- 5) 
2 • 2 


+ 3±v9 + 40 +3±7 

---, or-i- 


- . 3 + 7 10 5 , 

o. .. Xi — ^ ^ — 2> and X2 


nl = r J = _ 1 

4 4 



















QUADRATIC FUNCTIONS AND EQUATIONS 127 


139. The sum and the product of the roots of a quadratic 
equation have been used to check equations. We shall now 
prove the correctness of the rules used. The general quadratic 
equation is ax^ + 6 .r + c = 0 . 

In § 138 you proved that the roots of this equation are 

h — 's/b^ — 4: ac 


n = 


6 + V 6^ — 4 ac 


2 a 


and r 2 = 


2 a 


I. 


II. 


ri + r2 = 


— b+vb^ — 4 ac — b — — 4 

2 a 


, -2b b 


n • r2 = 


6 + V62 - 4 aA(- b - V62 - 4 




= 


2 a /\ 2 a 

{-by - {V¥ - 4 acy 
4 a? 


nn = 


- {b^ 


4a2 


4 ac) OH 
-or - 




Stated in words: for the quadratic ax^ -{■ bx 4- c = 0, the sum 

b . c 

the roots is -, and the product is — • 

a a 

Example. What are the sum and the product of the roots of 


2x^ = 9 X + 5? 

Solution. 1. 2 = 9 X + 5. 

2. 

to 

1 

CO 

1 

Oi 

II 

p 

3. 

.*. a = 2; b = — 9; c = 

4. 

.■.n + r2= -\= 

5. 

c - 5 i 

'• = a = : 

Note. 

Always arrange the quadratic as in 


the left side, and make the coefficient of positive. 


















128 


ALGEBRA 


140. The example at the bottom of page 126 shows how im¬ 
portant the quadratic formula is. You can and should memorize 
the general equation ax^ -{• hx c = 0 and the formula in two 
minutes. 

Notice that we were able to get this formula because we know 
how to "'complete the square/’ so that method also is important. 
However, from now on, when you cannot solve a quadratic 
quickly by factoring, solve it at once by the formula. 

Remember that there are two roots. Pattern your solution 
after that at the bottom of page 126. 


Find the roots correct to 
elusion stated on page 127. 

1. a:2+7a; + 6 = 0 

2. - 2x - = 0 

3. 2x2-5a: + 2 = 0 

4. 3 + 7 2/ + 2 = 0 

6 . 2 - 72 + 3 = 0 

6. 2w2_3^_2 = 0 

7. + 2 a: - 35 = 0 

8. 12 - 5 a; - 2 = 0 

9. 10m2 + 11 m - 6 = 0 

10 . 24 a;2 - 14 a: - 3 = 0 

11. 6 m2 = 7 w + 3 

12. 8 a:2 + 2 a; = 1 

13. 2 ^2 — ^ 

14. 4 — a: = 5 a:2 

16. 0 = 3 y2 _ 20 - 7 y 

16. a;2 + 2 a: - 2 = 0 

17. a:2 - 2 a: - 1 = 0 


RCISE 76 

tenths; check by using the con- 

18. 2 a:2 - 3 a: - 1 = 0 

19. 3 2/2 — 4^=! 

20. 2^2 = 82 + 1 

21 . 4 / 2+92 + 2 = 0 

22 . 5 c2 - 4 c = 2 

23. 3 a2 = 4 _ 6 a 

24. 3 c2 - 2 = 7 c 

26. a; + 5 = 3 a;2 

26. a^2 _ 5 ^ = 5 

27. a^2 _ ,7 ^ _ 3 = 0 

28. i/2 + .6 y - .4 = 0 

29. s2 - 2.5 2 - 1.5 = 0 

30. w 2 + 3 ^ _ 0 3 = 0 

31. 2 a:2 - 3 ar -f d = 0 

32. 3 a:2 + ca: + 5 = 0 

33. rx^ — 2 X t = 0 

34. 2 mx^ + 3a; — 4a=0 


QUADRATIC FUNCTIONS AND EQUATIONS 129 


141. When solving fractional equations we often get a 
quadratic. A difficulty, often met, is illustrated by the follow¬ 
ing example. 

T7' 7 C! 1 ^ ^ I ^ ^ ^ rv 

Example. Solve -- + -5 -h — = 0 

— I x^ — X x^ X 

After solving this equation it appears that x = 1, or — f. 

4 — 2 

When 1 is substituted for x, does — I -1- # = 0? 

0 0^ 

4 — 2 

But - j and do not have any meaning. (§ 20.) From this 

we conclude that 1 cannot be a root of the given equation. 

The value — f does not cause this same trouble. 

So, when solving a fractional equation, an apparent root must 
he rejected if it makes a denominator haw the value zero. 


EXERCISE 77 

Express surd roots correct to tenths. 


1. A - - - 5 = 0 

5 x^ X 


2 -_-— 

' X 2 a: + 3 


= 3 


6 . 


y + 8 

y - 2 

2 

1 

t - 1 

3« + 1 

2 

1 

2-4 

2-2 

7 

a: — 1 


= 2 

1 
2 


2a: — 3 a: — 2”*~2 


7. c - 


3 

2c - 3 


3r+ 1 3r 


+ 7^ = - 2 


9. 

10 . 


+ 3 6 ^ 

c 1 


11 . 1 + 

12 . 

13. 

14. 

15. 


2c+ 5 2c - 3 
2 3 


= 1 


2p - 3 p - 4: 
m m 


2m+l 3m— 1 
2 1 3 


= 1 


l-4y 5 l + 2y 
2 a: — 1 _ 2 a; + 1 _ ^ 
X + 1 X - 1 
9w — 2 w — 1 


16. 


5 '7C + 1 

14 — 3 m 6 + m 


4 — m 


= 1 
= 5 































130 


ALGEBRA 


EXERCISE 78 

1 . If 24 be added to 5 times a certain number, the result 
equals the square of the number. What is the number? 

2 . If twice the square of a certain number be diminished by 
the number itself, the result is 15. What is the number ? 

3. There are two consecutive even integers whose product is 
624. What are they? 

4. There are three consecutive odd integers such that the 
square of the first, increased by the product of the other two, 
is 844. What are they ? 

6 . The sum of a certain number and its reciprocal is 
What is the number ? (Hint. The reciprocal of x is -.) 

X 

6. The denominator of a certain fraction exceeds its numera¬ 
tor by 5. The sum of the fraction and its reciprocal is 
What is the fraction ? 

7. The base of a certain triangle exceeds its altitude by 7 
inches. Its area is 99 square inches. What are its dimensions ? 

8 . The area of a certain rectangle is 405 square feet. The 
sum of its base and altitude is 42 feet. What are its dimensions ? 

9. The total area of certain two squares is 765 square inches. 
The side of one exceeds the side of the other by 3 inches. How 
long is the side of each ? 

10 . The area of the main waiting room of the Union Railway 
Station at Washington, D. C., is 28,600 square feet. The sum 
of its length and width is 350 feet. What are its dimensions ? 

11 . The altitude of a certain right triangle is 6 feet more than 
its base. The hypotenuse is 30 feet. What are its base and 
altitude ? 

12 . If a certain number be increased by twice its reciprocal, 
the sum is 3^. What is the number ? 


QUADRATIC FUNCTIONS AND EQUATIONS 131 


13. A lawn is 30 by 40 feet. How wide a strip must be cut 
around it when mowing the grass to have cut half of it ? 

Hint. Referring to the figure, it is clear that 
]i w = the number of feet in the width of the 
border cut, then the dimensions of the uncut part 
of the lawn are (30 — 2 w) and (40 — 2 w). 

Hence, (30 - 2 iy)(40 - 2 ly) = I • 30 • 40. 

2(15 - WJ) • 2 • (20 - w;) = i • 30 • 40. 

14. A boy is plowing a field whose dimensions are 20 rods 
and 40 rods. How wide a border must be cut around it in order 
to have completed f of his plowing ? 

16. The numerator of a certain fraction is 3 less than its de¬ 
nominator. If 4 be added to both numerator and denominator, 
the new fraction exceeds the old by -J. What is the fraction ? 

16. A man traveled 75 miles by automobile at a certain rate. 
By increasing his average rate 5 miles per hour, he made the 
return trip in f hour less time than the time going. What was 
the rate going and returning ? 

17. Suppose a motor boat, traveling at the rate of 12 miles 
per hour in still water, goes 30 miles downstream and back 
again in a total of 5-|- hours. What is the rate of the current of 
the stream to the nearest tenth of a mile ? 

18. A crew can row 8 miles downstream, and back again, in 
4f hours. If the rate of the stream is 4 miles per hour, what is 
the rate of the crew in still water ? 

19. The denominator of a certain fraction exceeds its numera¬ 
tor by 3 . If the numerator be trebled, and the denominator be 
doubled, the sum of the resulting fraction and the original 
fraction is 1. What is the fraction? 

20. Some boys are canoeing on a river, in part of which the 
current is 5 miles an hour, and in another part 3 miles an hour. 
If, when going downstream, they go 4 miles where the current 
is rapid and 8 miles where it is less rapid in a total time of If 
hours, what is their rate in still water ? 

Note. Additional problems appear on Page 248. 


w [w 

w]^ 

• 40-2w 

1 

1 

!? 

1 

1 

[CN 

1 


1 

jco 

1 

yt} ___« • * «— 

1 

l__ 

Iw — 

"w"! w’ 





132 


ALGEBRA 


Imaginary Numbers and Roots of a Quadratic 

142. A new difficulty and a new kind of number. 

Example 1 . Solve the equation — 2 x 5 = 0. 
Solution. 1. Use the formula method of solution. 

a = 1; b = - 2; c = 5. 


- (-2) ± VT 

2 


2)2 - 4 • 1 • 5 


1 


2 -t V4 - 20 _ 2 db V- 16 


What does V 
4 is not the square root of 


16 mean? 

_16, since (— 4)2 = + 16. 

+ 4 is not the square root of — 16, since (+ 4)2 = +16. 

No number, studied so far, will produce a negative result when it 
is squared, so we have met a new difficulty. Let us try to solve the 
equation graphically. 


When X 
is 

Then y 
is 

- 3 

+ 20 

- 2 

+ 13 

- 1 

+ 8 

0 

+ 5 

+ 1 

+ 4 

+ 3 

+ 8 

+ 2 

+ 5 

+ 4 

+ 13 

+ 5 

+ 20 



According to the rule on page 118, the roots of the equation 
are the abscissas of the points where the parabola crosses the 
a:-axis. But this parabola does not cross the x-axis. This con¬ 
firms our realization that a new difficulty is introduced by this 
equation. 

The manner in which mathematicians finally overcame this 
difficulty is taught on pages 133 to 136. 
































































QUADRATIC FUNCTIONS AND EQUATIONS 133 


143. Mathematicians settled this difficulty by inventing a 
new kind of number — an imaginary number they called it. 

An imaginary number is the indicated square root of a nega¬ 
tive number, as V— 6; V— 3; V— 

The numbers studied previously, in contrast to the imagi¬ 
nary numbers, are called real numbers. 

Note. Negative numbers were once called numerae fictae, — fic¬ 
titious numbers. Both kinds are now very real to all scientists. 


144. Every imaginary number can be expressed as the 
product of a real number and V— 1. 

V— 1 is indicated by i, and is called the imaginary unit. 


Thus, V- 16 = Vl6(- 1 ) = 4 • V- 1 = 4^ 

V — = Va2(— 1 ) = Va^ • V— 1 = ai 

vCr~5 = VK{- 1) = ^ V5 = iV5 





5 ± 3 ^ Vs 
2 


The numerator of this result is a complex number. 


EXERCISE 79 


Express in terms 

of 

the unit i 







1. 



7. 


1L2 


13. 



19. V 

_ 2 

S 

2. 



8. 

V- 

27 


14. 

V- 

~ iV 

20. V 

_ 4 
¥ 

3. 

V- 169 


9. 

V- 

32 


16. 

V- 

“ ss 

21. aV 

_ 1 

■6 

4. 

V- 81 


10. 

V- 

63 


16. 


5 

■9 

22. 

_ 3 

8 

6. 

V- 64 


11. 

V- 

48 


17. 

V- 

_ 3 

T-B- 

23. V 

/- 

~ A 

6. 

V- 196 


12. 

V- 

75 


18. 

V- 

_ 7 

T 

24. V 

“ A 

26. 

3 ± 

5 

■9 

28. - 

f± 

V 

9 

Te" 

31. - 

f ± 

A 

26. 


f 

29. - 

f± 

V 

_ 27 

ST 

32. -^zLV— 

27. 

f±v- 

A 

30. - 

f± 

V 

~ 1 


33 . 1 


































134 


ALGEBRA 


1 = 8 • 1, or 8 , so 3 
(3-4i) = 2 + 5z- 


■ -i + 5 ^ = 8 
3 + 4^ = 92- 


1 . 


145. * Adding and subtracting imaginary and complex 
numbers. 

Just as 3 • 1 + 5 
similarly, (2 + 5 i) - 

146. Powers of the imaginary unit i. 

Since i = V— 1, then i^=- 

• i = — 1 ‘ i. 

i)(- 1) = 1. 

Similarly: = i; = — 1; = — i; i 


1 . 


_ 

= 1 . 


= + L 


^ Ky 

147. Multiplication of imaginary and complex numbers. 

Example 1. V— 2 • V— 3 = za/2 • ^V3 = = — 6. 

Example 2. Find (2 — V— 3) (5 + 3). 

Solution. 1. (2 - V^)(5 + V^) = (2 - iV3)(5 + zVJ) 

2. =10-3 iVs - ^•2 . 3 

3. = 10 - 3 2:V3 - (- 1) . 

4. =13-3 iVs. 

EXERCISE 80 

Find the following products: 


. V- 16 

• V- 18 

• V- 10 



V — 2b x^ • V — 4 
3V^ . sV^ 

7. W~^ • eV^ 

16. (- 2 + 2 

16. (6 - \/^)2 18. Hi- 1 + V^)12 

17. (a + 19. {i(- 1 - V^)) 

* Note to the Teacher. Pages 134-135 can be omitted. 


2) (3 - V^) 

6 ) (2 + V~=^) 
(2 + V ^)(2 + 3 V ^) 

(8 - V ^)(10 + V^) 
V^) 


2 








QUADRATIC FUNCTIONS AND EQUATIONS 135 


148. Division of imaginary and complex numbers. 

10 _ _ 10 ■ iV2 ^ 10 ^ StVi 

iV2 ■ iV2 ■ iV2 2 - 1 

To rationalize a denominator multiply it by itself. 

EXERCISE 81 

Express the quotient with rational denominators. 
V- 36 V- 40 a; 


= - biV2 


2 . 


V- 6 
V- 32 


3. 


4. 


V— 5 X 
6 


5. 


8 V- 2 

9. Illustratwe example. 


15 

14 


7. 


8 . 


VT6 

V28~ 

V- 14 


1 + V- 3 
2(1 - iVz) 2(1 


1 + iVs 
iVz) 


etc. 


(l + iV3)(l -'iV3) 1-3^2 

Note. (1 — i^) and (1 U^Vs) are conjugate complex numbers. 
To rationalize a complex denominator, multiply the terms of 
the fraction by the conjugate of the denominator. 

3 - V~^ 


10 . 


11 . 


1 + 

13 
3 - 

4 


2 - 


13. 


14. 


16. 


1 + V- 

1 - \/^ 


1 + V- 1 


12 . 

19. Find the value of + x 
X + 1 


16. 


17. 


18. 


3 + 2V- 2 

5 + 

2 _ 

4+ 


20. Find the value of 


5 - 2V^ 
— 1 when X = 1 + V— 2 

when X = 2 + V— 3 


21. Find the value of x^ — 2 x + 3 when x — 1 V 2 


22. Find the value of 


3 — X 


when X = 3 — V— 5 


































136 


ALGEBRA 


149. Solving quadratics which have imaginary roots. 

Example. Solve the equation + x + 2 = 0. 

Solution. 1. a = 1; 6 = 1; c = 2. 

- 1 ± -1 =h -1 ± ^V7 

2. :. X - 2 ~ 2 2 * 


3. 


- 1 + iV7 

:. xi =-2-i ^2 


- 1 - iV? 


2 


We shall also solve this equation graphically. 
Solution. 1. Let ?/ = + x + 2. 


When X = 

0 

+ 1 

+ 2 

+ 3 

- 1 

- 2 

- 3 

- 4 

then y = 

+ 2 

+ 4 

+ 8 

+ 14 

+ 2 

+ 4 

+ 8 

+ 14 


3. The graph has the same 
shape as the graphs obtained 
when solving other quadratic 
equations; but the graph does 
not cross the horizontal axis at 
all. Hence, ^ or x^ + x + 2 is ' 
never zero for any real value 
of X. 

Remember this is char¬ 
acteristic of the graph of a 
quadratic which has imagi¬ 
nary roots. 



1 





Vl 








\ - 







t 





T" 




5 



i : 





\ 






i 

i 





-^r 






1 







\ - 


-•+1 

t) - 


-t- 







‘V 





? ■ 







\ 




i 








s 



c.. 

j 









s 


J 

/ 












/ 









's 

S - 



















> 

( 





rr 





Xi 




- 





►2 1 

t3 ^ 

■4 ^ 







































EXERCISE 82 

Solve the following equations either by completing the square 
or by the formula. Draw the graphs of the first three equations. 


1. + a: + 3 = 0 

2. x2 + 2x+2=0 

3. x2-3x-i-4=0 

4. 2x^ — x-f-l=0 

6. 3y2_22/+2=0 


6. 2a2-3a + 2= 0 

7. 3<2-4f + 2= 0 

8. 4r2-5r+2=0 

9 . 7 5 ^-|- 65 ~ 1“3 = 0 

10. 6 c2 - 5 c 4- 2 = 0 


















































































QUADRATIC FUNCTIONS AND EQUATIONS 137 


150. The numbers considered in this text so far are: 

I. The real numbers, which include: 

(a) the rational numbers, which are the positive and nega¬ 

tive integers, and the fractions whose terms are such 
integers; 

(b) the irrational numbers, which, thus far, are the quad¬ 

ratic surds and surd expressions. (See p. 105.) 

II. The imaginary and complex numbers. (See p. 133.) 

Historical Note. Greek mathematicians as early as Euclid were 
able to solve certain quadratics by a geometric method, about which the 
student may learn when he studies plane geometry. Heron of Alexan¬ 
dria, about 110 B.C., proposed a problem which leads to a quadratic. 
His solution is not given, but his result would indicate that he probably 
solved the equation by a rule which might be obtained from the quad¬ 
ratic by completing its square in a certain manner. Diophantus, 
275 A.D., gave many problems which lead to quadratic equations. 
He considered three separate kinds of quadratics. He gave only one 
root for a quadratic, even when the equation had two roots. 

The Hindu mathematicians, knowing about negative numbers, con¬ 
sidered one general quadratic. The Hindus knew that a quadratic has 
two roots, but they usually rejected any negative roots. 

The Arabians went back to the practice of Diophantus in considering 
three or more kinds of quadratics. Mohammed Ben Musa, 820 a.d., had 
five kinds. He admitted two roots when both were positive. Alkarchi 
gave a purely algebraic solution of a quadratic by completing the square, 
and refers to this method as being a diophantic method. 

In Europe, mathematicians followed the practice of the Arabians, and 
by the time of Widmann, 1489, had twenty-four special forms of equa¬ 
tions. These were solved by rules which were learned and used in a 
mechanical manner. Stifel, 1486-1567, finally brought the study of 
quadratics back to the point that had been reached by the Hindus one 
thousand years before. He gave only three normal forms for the quad¬ 
ratic; he allowed double roots when they were both positive. Stevin, 
1548-1620, went still farther. He gave only one normal form for the 
general quadratic, as do we; he solved this in both a geometric and an 
algebraic manner, giving the method of completing the square. He 
allowed negative roots. 


138 


ALGEBRA 


151. The different kinds of roots of a quadratic. 

Consider the following equations, their roots and graphs. 


(a) 2 X — 8 = 0 

- 2 ± V36 - 2 ± 6 

^ “ 2 2 

xi = — 4; 3^2 = 4~ 2 

(b) x^+2x-2 = 0 

- 2 ± Vl2 - 2 =b 2 Vs 
^ " 2 " 2 

ici = — 1 — Vs, or — 1.7 
a ;2 = ~ 1 4“ Vs, or + .7 

(c) a:2 + 2 X + 1 = 0 


- 2 ± V4 - 4 



xi = - - -/ or - 1 

(d) x2 + 2x + 5 = 0 

- 2dz V- 16 -2zh4:i 

- o ~ 2 


- 24-0 


X2 = 

2 

or — 1 

Xl = 

1 

to 

1 

cS>. 

> or — 1 — 2 2 

2 

X2 = 

-2 4-42 

-or — 14-22 

2 


In 

The roots are 

62 — 4 ac is 

The graph 

Equation (a) 

Real 

Rational 

Unequal 

Positive and a 
perfect square 

Crosses the x-axis 
at two points 

Equation (6) 

Real 

Irrational 

Unequal 

Positive and not 
a perfect square 

Crosses the x-axis 
at two points 

Equation (c) 

Real 

Rational 

Equal 

Zero 

Touches the x-axis 
at a point 

Equation (d) 

Imaginary 

Unequal 

Negative 

Is entirely above 
the x-axis 













































































































QUADRATIC FUNCTIONS AND EQUATIONS 139 


152. It is possible to determine the kind of roots of a quad¬ 
ratic equation (with rational coefficients) without first finding 
the roots. 

The roots of + 6a: + c = 0 are, as you know, 

— 6 + V62 — 4 ac , — 6 + a/ 62 — 4 ac 

n = -- and r 2 = -r- 

2 a 2 a 

The kind of roots is determined by the value of 6^ — 4 ac 
as you have seen on page 138. For this reason, 6^ — 4 ac is 
called the discriminant of the quadratic. 

Rules. 1. If — 4: ac is 'positive^ the roots are real and 
unequal. They are rational if b^ — 4 ac is a perfect square; they 
are irrational if b"^ — 4 ac is positive and not a perfect square. 

2. If b^ — 4 ac = zero, the roots are real, rational, and equal. 

3. 7/62 — 4 ac is negative, the roots are imaginary or complex 
numbers. 

Example 1. By inspection, determine the kind of roots of 
2a:2 - 5a: - 18 = 0. 


Solution. 1. b^ — 4 ac ={— — 4(2)(— 18). 

2. 62 _ 4 ac = 25 + 144 = 169, or IS^. 

3. by Rule 1, the roots are real, rational, and unequal. 

Example 2. What kind of roots has 3a:2 + 2a:+l = 0? 
Solution. 1. 62 — 4 ac = (2)2 — 4(3)(1), or — 8. 

2. .•. by Rule 3, the roots are complex numbers. 


EXERCISE 83 

the kind of roots of: 

7. 3 a: - 9 a:2 + 1 = 0 


Determine by inspection 

1 . 2/2 _ 7 ^ + 12 = 0 

2. 9 a:2 - 24 a: + 16 = 0 

3. 2m2 — m+l = 0 

4. 3 2/2 = 5 2 / + 2 
6. 7 a:2 + 5 a: = 9 

6. 1 + 6 i + 4 f2 = 0 


8. a:2 = 2 a: — 7 

9. 10 = 4 a: + 5 a:2 

10. 16 m2 - 4 = 0 

11. 9 a:2 - 30 a; + 25 = 0 

12. a:2 - 4 a: + 8 = 0 






140 


ALGEBRA 


EXERCISE 84 

Find the sum and the product of the roots of the following 
equations without solving the equations. (See p. 127) 


6 . 4 a = 7 — 3 
6 . + bhx — — 


1. a^2 + 8x+7 = 0 

2 . _ 9 2 + 14 = 0 


3. 8 — 2 X — 15 = 0 7. 15 = 2 ax 

4. 63 ?/2 - 16 ?/ + 1 = 0 8 . 3 x2 = 4 x^ + 2 

9. One root of 3 x^ — 2 x — 5 = 0 is — 1 . Find the other 
root. Suggestion. What is the sum of the roots? 

10 . One root of2x2 — 7x+3 = 0is3. Find the other. 

11 . One root of 6 x^ — 1 = 5 x is — Find the other root. 

12 . Find k so that one root ofx^ + 4x — ^ = 0 will be — 12. 

Solution. 1. ri + r 2 = — 4. — 12 + r ’2 = — 4. .*. r 2 = 8. 

2. But nr. = - k. - k = {- 12) -8; :. k = 1 

Check. Substitute in -j- 4 x — A: = 0 the value of k you find; 
solve the equation, and thus determine whether — 12 really is one root. 

13. Find k so that — f may be a root of 6 x^ + 7 x + 2 A; = 0. 

14. Find k so that f may be a root of 8 x^ — 10 x = — k. 

15. Find a so that 7 may be a root of 7 2 ;^ — 48 2 — a = 0. 

16. Find n so that f will be a root of 3 x^ + nx = 4. 

17. Find p so that — f will be a root of 10 x^ — 5 = px. 

18. Find c so that the roots of2 22 — 82 + c = 0 shall be 

equal. 

Solution. 1. Let the two equal roots be represented by r. 

2. .•. r + r = ? .•. r = ? But r • r = c = ? 

Or the example can be solved by using Rule 2, § 152. 

Find: 

19. t so that the roots of3x2 — 4x+A = 0 shall be equal. 

20 . s so that the roots of 4 x^ — 5 X + 9 — 0 shall be equal. 

Note. Additional examples appear on page 249. 


QUADRATIC FUNCTIONS AND EQUATIONS 141 


153. Forming quadratic equations having given roots. There 
are two methods. 

Example 1. Form the equation whose roots are ^ and — 
Solution. 1. lix = then x - ^ = 0. 

2- If a: = — f, then a; + f = 0. 

(aJ - D(a: + I) = 0, or a:2 + 1 a: - I = 0. 

4. 8 a:2 + 2 a: - 3 = 0. 

Check by solving the equation. 

Note. This method may be used when forming an equation which 
shall have three or more given roots. Use this method in the even 
examples below. 

Example 2. Form the equation whose roots are 2.4 and 

-.5. 


Solution. 1. Let the coeflEicient of a;^ be 1. Then the equation is 
— {n + r 2 )x + rir 2 = 0. (See § 139) 

2. But n + r 2 = 2.4 + (- .5), or 1.9; 

and n - r 2 =■ 2.4 • (— .5), or — 1.2. 

3. .'. the equation is a:^ — 1.9 a; — 1.2 = 0. 

Check by solving the equation. 

EXERCISE 85 


Form the equation whose roots shall be 


1. 2, 5 

2. - 3, 1 


4. 

6. 

6 . 

7. 

22 . 

23. 


6 , - 1 

-3,-7 

if 
f» ~ i 

1.5, 2 

1 + V2 1 - V2 

2 ' 2 

2 + V3 2 - \/3 

4 ' 4 


8. .2, .5 

9. 1.5, - 2.4 

10. - 3.6, - .25 

11. “h 3 771, “h 2 777 

12. - 4 /, + 7 ^ 

13. i c, - f c 

14. d e, d — e 


24. 


1 + 27 1 — 27 
2 2 


3-73 + 
3 ’ 


16. V5, - Vb 

16. 1 + V2, 1 - V2 

17. V3 + 2, V3 - 2 

18. 3 + Ve, 3 - V6 

19. 6 + 2 c, 6 - 2 c 

20. 'Vm, — \/m 

21. a + Vh, a — V6 

a — a + Vb 
c c 

„ a — ib a + ib 

27. -j- 


c 


c 














142 


ALGEBRA 


EXERCISE 86. CHAPTER MASTERY-TEST 

1. The function 3 - 2 x - 5 is of the__degree. 

2. (a) Draw the graph of the function 3 x^ - 2 x — 5. 

(6) From the graph give the roots of the equation 

3x2-2x-5 = 0. 

3. Solve the equation 2x2-5x + 3 = 0by factoring. 

4. Solve the equation x2-4x-5 = 0by completing the 
square. 

6. Solve the equation 2 x^ - 3 x - 4 = 0 by the formula. 

6. Without solving the equation 3x2 — 4x — 2 = q. 

(a) Tell the sum, the product, and kind of its roots; 

(b) Does the graph of 3 x^ — 4 x — 2 cut the x-axis? 

7. Form the equation whose roots are 1 + ^3 and 1 — \^ 3 . 

8. Tell the sum, the product, and the kind of roots of 

(a) 2 x2 — 3 X + 4 = 0, without solving the equation. 

(b) Does the graph of 2 x^ — 3 x + 4 cut the x-axis? 

9. The graph of every second degree function of one variable 

is a_ 

10. (a) Tell the sum, the product, and the kind of roots of 
the equation 4x2— 12x4-9 = 0. 

(b) Does the graph of4x2— 12x4-9 cut the x-axis? 

11. If a man travels 120 miles by auto and returns at a rate 
which is 10 miles an hour more, he wiU require 7 hours for the 
whole trip. What was his rate each way? 

12. A boy was mowing a lawn, 40 ft. wide and 90 ft. long. 
How wide a border must he cut around its outside lines to have 
completed five sixths of his work? 

13. S = at + i gf. Express t as a function of a, S, and g. 

14. One root of2x2 — 15x4-A; = Ois twice the other. Find 
the value of k. 




QUADRATIC FUNCTIONS AND EQUATIONS 143 


EXERCISE 87. CUMULATIVE REVIEW 


1. Find the value of 


2 — Z X — ^ 

2^3 _ 1 


when a: = f. 


2. A boy is plowing a field which is 30 rods wide and 50 rods 
long. How wide a strip must he plow around the outside so 
that 2 acres (320 sq. rd.) shall remain to be plowed? 

3. A man agreed to work at a certain job on condition that 
he receive $9 for each day that he worked. His employer agreed 
on the condition that the man forfeit $5 for each day that he 
failed to appear until the job was finished. At the end of 
40 days (excluding Sundays) he received $220. How many 
days did he work? 


4. Rationalize the denominator and find the value to hun¬ 
dredths : 


(a) of 


1 + V5 

V2 


{h) of 


V5 - a/2 

Vs + V2 


6. Write as a single fraction in lowest terms. Check, letting 
a = 2. a a 

CL 2 CL — 3 


CL 2 a — 3 


6. Solve the equation — 2 6a; + c = 0 by completing the 
square. 

7. Find X to the nearest hundredth: x^ — .2 x — 2.05 = 0. 

8. A and B working together can do a piece of work in 
6 days; they can also do the work if A works 10 days and 
B works 3 days. How many days would it take each to do 
the work alone? 

9. If a certain number be divided by the sum of its digits, 
the quotient is 5 and the remainder is 2. If the digits be re¬ 
versed, the sum of the resulting number and 58 is twice the 
given number. Find the number. 










144 


ALGEBRA 


10. Represent graphically the function 2 x. From this 
graph, determine the roots of the equation x^ -\- 2 x = 5 to the 
nearest tenth. 

11. A fast train runs 8 miles per hour faster than a certain 
slower train. It requires 2 hours less time for a trip of 140 miles 
than does the slower train. Find the rate of each train. 


12. A crew rowed 12 miles downstream, and back again in a 
total time of 8 hours. The rate of the current is known to be 
2 miles per hour. At what rate did the crew row? 


13. One machine can do a certain piece of work in 4 hours 
less time than another. When working together, they do the 
work in 1§ hours time. In how many hours can each do it alone? 

14. A farmer is plowing a field whose dimensions are 40 rods 
and 90 rods. How wide a border must he plow around the field 
in order to have completed § his plowing? 

a — Vb , a + Vb 


15. Form the equation whose roots are 

16. Find the prime factors of: 


and 


(a) x‘^°' — 12a:“ + 32 

(b) ax^ — 16 a 


(c) x^ — 64 




(3 X - 2)= 


17. Determine the roots ofx^ + 4x — 5 = 0 graphically. 

18. Find the number of two digits such that, if the digits 
be reversed, the difference of the resulting number and the 
original number is 9, and their product is 736. 

19. A rectangular field contains 2^ acres. If its length be de¬ 
creased by 10 rods, and its width by 2 rods, its area would be less 
by one acre. Find its length and width. (lA. = 160 sq. rd.) 

20. The speed of an airplane is 90 miles an hour in calm 
weather. Flying with the wind, it can cover a certain distance 
in 4 hours but, when flying against the wind, it can cover only 
f of this distance in the same time. What is the velocity of 
the wind? 




IX. GRAPHS OF EQUATIONS OF THE SECOND 
DEGREE 

Two Variables 

154. You should now learn the names of certain curves and 
their standard equations. 

155. The parabola. Simplest form of its equation: y = 


When X is 

When y is 

- 3 

+ 9 

- 2 

+ 4 

- 1 

+ 1 

0 

0 

+ 1 

+ 1 

+ 2 

+ 4 

+ 3 

+ 9 


This graph is drawn on axes 
which have the same scale on 
both the a:-axis and the 2 /-axis. 

Variations of this simple equation produce easily understood 
changes in this standard graph. 

Example. Consider y = ^ x‘^. Here the ordinate is J the 
square of the abscissa. We can get the graph from that above 
by taking one-half of each ordinate, without computing the new 
ordinates arithmetically. The resulting graph is shown by a 
dotted curve. (Thus, above, when x = 2, y = 4:. In this 
example, when x = 2, y — ^{4), or 2.) 

We shall call this sketching the graph. 

145 






































































146 


ALGEBRA 


156. Many laws of nature and relations are pictured by this 
parabola or modifications of it. (Continued from p. 145.) 

Example. *S = | gf, ov S = (about) 16 represents the dis¬ 
tance through which an object will fall in t seconds. 

S changes as the square of t changes. 

If this graph is drawn on the small scale shown on page 145, the 
parabola becomes very narrow, as each ordinate for this graph must 
be 16 times as long as the corresponding ordinates on page 145. 

EXERCISE 88 

1. On a large sheet of graph paper draw the graph oi y = 
for values of x from — 5 to +5. {Suggestion. If you use 
paper like that used for the graph on page 14fi, take one large 
space for the unit, both vertically and horizontally.) 

2. On the same sheet as used for Example 1, sketch the graph 
of y = J x^. (Use a dotted line. Do not compute ordinates. 
Rather make each ordinate f as long as the corresponding 
ordinate in Example 1.) 

3. On the same sheet sketch the graph of y = 1.5 (Use 
a dash line. See Example 2.) 

4. A = TTr^ is the area of a circle. Let tt = 3.1. Then 
A = 3.1 r^. Draw this graph as follows: 

(a) On a large sheet, first draw with a dotted line the graph 
of A = r^. 

ih) Then sketch the graph of A = 3.1 A, remembering that 
each ordinate for it will be a little over 3 times the corresponding 
ordinate of A = r^. 

5 . S = TTr^ is the formula for the surface of a sphere. 

{a) What will be the shape of the graph of this equation? 

(b) Tell how to sketch it, following a plan like that for 
Example 4. 

6. Can you sketch the graph of y = x"^ + 5 without com¬ 
puting values of 


GRAPHS OF EQUATIONS OF SECOND DEGREE 147 


157. The circle. Simple form of its equation: = r^, 

where r is the radius and the center is at the origin. 

Example 1. Draw the graph of — 25. 

Solution. 1. 2/2 = 25 — or ?/ = zhV^25 — x^. 

2. When x = — S, y = ±V25 — 9 = ±Vl6 = dz4. 

Therefore when x = — S, y = + A; when x = — S, y = — 4:. 

3. In this manner we obtain the table. 


When X = 0 

± 1 

±2 

± 3 

± 4 

±5 

then 2 / = ± 5 

=fc 4.8 

±4.5 

± 4 

±3 

0 


This means: when x = -{- 1, 

2 / = + 4.8; when x = — l^y = 

+ 4.8; when x = + 1, y = 

— 4.8; when x = — 1, y = 

— 4.8, etc. 

4. When x is greater than 5, 
y is imaginary. 

Thus, when x = 6, y = 

± V25 - 36 or =b This 

means that there are not any 
points on the graph for values 
of X greater than + 5. Neither 
are there for values of x less 
than — 5. 

5. The graph appears at the right. The graph is a circle if the unit 
is of the same length on both the x-axis and the 2 /-axis. 

EXERCISE 89 

1. What is the graph (or locus) of + 2 /^ = 81? Draw it 
without computing ordinates. 

2 . On the same set of axes sketch the graph of + 2/^ = 100. 

3. On the same set of axes sketch the graph of + 2 /^ = 50. 

4. (a) Sketch the graph of x^ + 2 /^ = 36 on axes which have 
units of the same length. 

(6) Now sketch it on a set of axes, on which the unit for y 
is made one half as long as the unit for x. 






















































































148 


ALGEBRA 


^2 y2 

158. The ellipse. Simple form of its equation: ^ ^ = 1 

Example 1. Draw the graph ^ + "g ^ 

Solution. 1. Solving for y, you would get ^ = ± 4 ”^225 — 9 x-. 

2. If a: = 2, 2 / = ± 2.7; so for each x, there are two values of y. 

3. Also \i X = — 2, you would get ^ = db 2.7, since (— 2Y = 4. 

4. When x is more than 5, y is imaginary; also when x is less than — 5. 


When X = 

0 

±1 

± 2 

± 3 

± 4 

±5 

over + 5 

Then y = 

± 3 

±2.9 

±2.7 

± 2.4 

± 1.8 

0 

imaginary 



5. The graph is an ellipse provided the units on the y-axis and on the 
x-axis are of the same length. 

The greatest width in the a’-direction is 10, just twice the 
square root of 25, the denominator of x^. This is called the 
major axis. The greatest width in the ^-direction is 6, just 
twice the square root of 9, the denominator of This is 
called the minor axis. 

Example 2. What is the graph oi x^ 4 y^ = 64? 

Solution. 1 . D64 _i_ y‘‘ — ^ 

M 16 “ 

2. The graph is an ellipse. Major axis = 16; minor axis = 8. 










































































































GRAPHS OF EQUATIONS OF SECOND DEGREE 149 


159. The equilateral hyperbola. The equation in simple form 
is xy = ky where k is an arithmetical number. 

Example. Draw the graph of xy = 6. 


Q 

Solution. 1. y — - 

2 . 


When X = ^ 

1 

U 

23; 

4 

6 

12 

1 

2 

- 1 

- n 

- 2 

- 3 

- 4 

- 6 

- 12 

then ^ = 12 

6 

4 

CO 

li 

1 

1 

- 12 

- 6 

- 4 

- 3 

- 2 

- u 

- 1 

1 

2 


3. The curve has the shape shown in the graph when the unit on 
the x-axis and i/-axis is the same length. 

Notice that y becomes smaller as x becomes larger; etc. 



Example. You know the formula d = rt. Let d = 100. Then 
rt = 100. If the vertical axis is used for the values of r, and the hori¬ 
zontal axis for values of t, then the graph will be an equilateral hyper¬ 
bola. (Negative values of r and t do not have any meaning.) 

EXERCISE 90 

1. Draw the graph of A = hb, when A = 50. 

2. Draw the graph of *S = i Ch, when S = 50. 





















































































































150 


ALGEBRA 


160.* The general hyperbola. Simple form of the equation 


^ ^ = 1 

O 7 0 • 




62 


Example. Draw the graph ~ x 

y ^ 


Solution. 1. Solving for y, you would get y = ± — 9. 

2. If X is less than 3 or greater than — 3, y is imaginary. 

3. If X = + 3, or — 3, y is zero. 

4. If X = + 4, you get y = ± 1.8. 

Also, if X = — 4, you get y — dz 1.8. 


.*. When X = 

0 

=b 1 

±3 

db 4 

d=5 

d= 6 

then y = 

imag¬ 

inary 

imag¬ 

inary 

0 

-1- 1.8 

-h 2.7 

4-3.5 



Y' 


6. The graph is a hyperbola. Its “center” is at the origin. Observe 
that the hyperbola consists of two pieces; that each half looks like a 
parabola, which draws closer and closer to the dotted lines. 

These dotted lines are easily located. Point B is the point (3, 2). 
Its abscissa is the square root of the denominator of x^; its ordinate is 
the square root of the denominator of y^. Similarly B' is the point 
(— 3, + 2). OB and OB' are called the asymptotes of the hyperbola. 

A is on the x-axis just below point B] A' is on the x-axis just below 
point B'. 

* This graph is not required by the C. E. E. B. 































































































X. SYSTEMS INVOLVING QUADRATICS 

Two Variables 


161. Systems consisting of one linear and one quadratic 
equation. 

Example. Is there a right triangle with hypotenuse 5 inches 
and altitude 1 inch more than its base? 


Graphical solution. 1. Let x = the base and y = the altitude. 


+ ^2 = 25 (1) 

y - X = \ (2) 

3. The graph of (1) is the circle with radius 5 inches. 

4. For (2): when x = 0, y = 1) 

when X = 2, y = 3. 

5. At A: a: = 3, ^ = 4. 

At J5: X = — 4:, y = — 3. 

Each is a solution of both equa¬ 
tions. 

(Only the values x = 3 and y = 4 
have any meaning for this problem, 
of course.) 

Algebraic solution. 1. From (2) 

y = X + 1. 

2. Substitute in (1). 

-j- + 2 a: + 1 = 25, 

or2x^-\- 2 a: — 24 = 0. 




3. .-. a;2 + a: - 12 = 0. 

4. .•. a: = — 4 1 

/. 2 / = — 4 + 1, or — 3 I 

Hence the common solutions are: 

Check: For a:= — 4, ^ = —3 
In (1): Does 16 + 9 = 25? A"es 
In (2): Does (-3) - (-4) = 1? Yes 


... (3; + 4)(a:-3) =0. 

and a; = + 3 
^ = 3 + 1, or 4 

X = 3, y = 4) 

X = - 4, y = - 

For X = 3, y = 4 
In (1): Does 16 + 9 = 25? Yes 
In (2): Does 4 — 3 = 1? Yes 


151 










































































152 


ALGEBRA 


162. A system consisting of one linear and one quadratic 
equation always has two solutions. 

Geometric explanation. The graph of the linear equation is 
a straight line. 

The graph of the quadratic equation is a circle, a parabola, an 
ellipse, or a hyperbola. 

Intersection points therefore occur in figures such as: 



The coordinates of A are a solution of the system; those of 
B are also a solution. Hence there are two solutions. 

If the straight line does not cut the curve, the solutions are 
imaginary. 


Example. 


Solve the system 


\Zc + 2d = - 2 
\ cd “h 8 c = 4 


( 1 ) 

( 2 ) 


Solution. 1. From (1), c = —- g - - ^ • (3) 

2. Substituting in (2), d ^j + 8 ^ ~ ^ ~ ^ ^ j = 4. (4) 

3. Simplifying, — 2 d — 2 _ 16 — 16 d — 12 = 0. (5) 

4. .*. 2 d2 + 18 d + 28 = 0, or d2 + 9 d + 14 = 0. 

5. Solving for d, d = — 2, or d = — 7. 



When d = — 2, 

When d = — 7, 

6 . 

in (1), 3c - 4 = - 2 

in (1), 3 c — 14 = — 2 


or c = |. 

or c = 4. 

7. 

.'. The solutions are A. { 



These solutions check when substituted in (1) end (2). 





























































































SYSTEMS INVOLVING QUADRATICS 


153 


1 . 


2 . 


3. 


4. 


6. 


6 . 


7. 


8 . 


10 . 


11 . 


12 


13. 


EXERCISE 91 

Solve the following systems: 

^2 2/2 = 85 

X - y = I 

4 + ^2 = 25 

2r + < = 7 
xy = 28 
X -\- y = ll 

a2 + 62 = 130 

a — h = S 
2r + t= 7 
rt = 6 

j x"^ — xy — y^ = 19 
\x - y = 7 
d — 2 c = S 
-3 cd = 22 
xy = 18 
y - X = 3 
x^ — xy y^ = 63 
X - y = - 3 
I ^2 _|_ 2/2 = 101 
\ a: + y = 9 
x^ y^ xy = 39 
X y = — 2 
I2y + 2x = 5 xy 
\2 X + 2 y = 5 
a:2 2/2 = 52 

X - y = 2 


14. 


16. 


16. 


17. 


18. 


19. 


20 . 


21 . 


22 . 


23. 


24. 


25. 


26. 


/ 2 a:2 + 2/2 = 54 

\x - \y = 9 
x"^ — 3 xy y^ = — 11 
y - X = - 1 
y - 3x = 7 
x'^ + xy = 2 

4:X^ — xy = 2(x + y) 
y - X = 1 

x^ y^ = a? 

2 X — y = a 
2/2 - a:2 = 20 
X = %y 

fa:2 + 2/2-62/ = 0 
\ y -h 2 X = 0 
4 3^2 2/2 = 40 

2 X -1- y = S 
X - y = 1 
xy = a? + a 
^2 4- 2/2 = 25 
X — ^y = 9 
\ x + y = - 3 
\xy = — 54 
[ a:2 + 2/^ = 16 — 6 a: 

\y = 2 X 1 
I x^ — xy = — 7 
\ 4 a: — 2/ = 0 


Solve Ex. 27 and 28 graphically and algebraically. 


27. 


a;2 4- 2/2 = 4 
a: 4- y = 4 


28. 


y — x^ = 9 
2 / — 4a:4-4 = 0 


154 


ALGEBRA 


163. Systems consisting of two quadratics. Read this and 
the following page as preparation for page 156. 


Example 1. 


Solve the system 


f + 2/' = 25 
1 + 2 2/2 = 34 


( 1 ) 

( 2 ) 


Graphical solution. 1. You 
know that the graph of (1) is a 
circle and of (2) is an ellipse. 

2. You know that the co¬ 
ordinates of the intersection 
points should be common solu¬ 
tions. 

At A: a;=-l-4, 2 / = 3 
AtB: a:= — 4, 2 / = 3 
At C: X = — 4, 2/ = — 3 
At D: x = + 4, 2/ = — 3 
These do satisfy both equa¬ 
tions when substituted. There 
are four solutions^ each con¬ 
sisting of real numbers. 



As you know, the graph of a quadratic equation having two 
variables is a circle, an ellipse, a parabola, or a hyperbola. The 
graphical solution of a system of two such quadratics will 
produce figures like the following: 



In no case are there more than four points of intersection, 
and hence four real common solutions. But these figures make 
it clear that there may be only two points of intersection, or 
even none at all. 





























































































































































SYSTEMS INVOLVING QUADRATICS 


155 


Example 2. 


Solve the system 


/ = 25 

U2-32/ + 3 = 0 


( 1 ) 

( 2 ) 


Solution. 1. Subtracting, y'^ S y — 2S = 0 

2. /. (2/ + 7)(i/-4) =0 

3. - 7; or?/ = + 4 

4. In(l); when?/=+4: 

= 9; or X = db 3. 

5. In (1): when ?/ = — 7: 

x^ = 25 — 49, or x^ = — 24 
X = rh V— 24, or X = ±2 iVq. 
There are four solutions: 


: (" = 
I y = 


C; 


+ 3 
+ 4 

x = + 2iV6 
y = - 7 


B: 


D: 



/ X = - 3 
l2/ = +4 


X = 

y = 


2?:V3 

7 


Note. The circle (x^ y^ = 25) cuts the parabola (x^ — 3 ?/ + 3 
= 0) at points A and B whose coordinates are solutions A and B. 

This figure makes it appear that two imaginary solutions C and D 
correspond to two missing points of intersection. 


Example 3. 


Solve the system 


^2 ^2 _ 25 

x^ — y = +5 


(1) 

( 2 ) 


Solution. 1. From (2): x^ = ^ + 5 
2 . ?/2 + ^ - 20 = 0 
3. .-.?/ = 4; ?/= - 5 

In (1): when y = — 5: x^ = 0, or x = (±)0. 
In (1): when ^ = + 4: x^ = 9, or x = zfc 3. 


The solutions are: 


A: 

fx = 0 

B 


1 y = - 5 


C: 


D 


t 2/ = + 4 



X = 0 
y = - 5 
/x = - 3 
1 2 / = + 4 



In the figure, points C and D correspond to solutions C and D; and 
point A must correspond to solutions A and B. 

Evidently, two identical solutions correspond to a point of tangency 
of the two graphs. 











































































156 


ALGEBRA 


164. A system consisting of two quadratics having two 
variables always has four common solutions; two or four of 
these may be imaginary or complex numbers; two or four may 
be identical real solutions. 

These solutions can be obtained algebraically for only certain 
systems. Description of the methods of solution is so lengthy 
that it is better to depend upon illustrative solutions. 

One method is illustrated in Examples 2 and 3 on page 155. 

Observe that one variable was eliminated, and then two values of 
the second variable were found. For each of these values of the second 
variable two values of the first variable were found, which, combined 
separately with the first variable, gave two solutions. Thus there 
resulted four solutions. 


A second method is illustrated by the solution below. 


Examine. Solve the system 


\ a? -E ah -\- IP' = 63 


62 = 


27 


( 1 ) 

( 2 ) 


Solution. 1. M 3 ( 1 ) 3 + 3 a6 + 3 6^ = 189 (3) 

2. M7(2) 7 a2 - 7 62 = - 189 (4) 

3. (3) + (4) 10 a2 + 3 a6 - 4 62 = 0 

4. .-. (5 a + 4 6)(2 a - 6) = 0. .*. (5 a + 4 6) = 0, and (2 a - 6) = 0. 

5. Now form and solve the following two systems: 


r a2 - 62 = - 27 
12a - 6 =0 
You will find the solutions: 


f a2 - 62 = - 27 
1 5a + 46 = 0 
You will find the solutions: 


A: when a = — 3, 6 = —6 
B: when a=+3, 6=+6 


C: when a = — 4x^3, 6 = + 5^3 
D: when a = + 4V^, 6 = — 5^3 


The plan in this solution is to combine the given equations so that a 
quadratic is secured which can be factored, and thus give two linear equa¬ 
tions. 


Note. Observe that the term not containing an unknown was 
eliminated. This method is suggested only when the sum of the exponents 
in each term which contains an unknown is two. 

Thus: in a2, the exponent is 2; in ab, the sum of the exponents of 
a and 6 is 2. 



SYSTEMS INVOLVING QUADRATICS 


157 


1 . 


2 . 


3. 


6 . 


6 . 


.7. 


10 . 


11 . 


2 — 3 = 19 

xy = — 6 

^2 _|_ 2^2 ^ 40 - 

xy = 12 
- ¥ = 3 
ah = — 2 

772,2 4 _ ^2 _ 20 

mn = — 5 
( x'^ + 3 xy = 28 
[xy + 4 y^ = 8 
x^ — xy = 4 
^2 4 _ ^2 _ 20 


19. 


20 . 


21 . 


22 . 


23. 


24. 


EXERCISE 92 

Solve the following systems of equations; 
r 2 ^2 - 7/2 = - 1 
■ + 2 7/2 = 22 

' 2 a2 + 62 = 7 
a2 - 2 62 = - 4 
f 4 — 3 7/^ = — 11 

[ 4 x 2 + 4 ^2 = 27 
i2x^ -3 y^ = - 6 
’ [ 4 x^ - y^ = 8 
f 4 a;2 + 7 7/2 = 32 
, 11 7/2 - 3 0^2 = 41 

' 2 r2 _ 5 ^2 23 

= — 3 


13. 


14. 


15. 


16. 


17. 


18. 


Solve the following systems for x and y : 


/ 3 cd + ^2 = — 14 

- cd = 30 

2x^ — 3 xy = 2 
4 + 9 7/2 = 10 

r 4 :r2 + 3 y2 = 28 
[2 x"^ -\- xy + y‘^ = 8 
/ 77 i 2 + mn = 8 
I mn — 7 i 2 = — 12 

/ a :2 + 7/2 = 37 

\ xy = 6 

— y‘^ = 19 

2 X + y2 = 5 

f 2 x 2 - xy = 2 

I 4 x2 + y2 = 10 

3 a 6 - 2 62 = 0 
2 o? — ah = 2 
7 i 2 + 3 mn == 2 
9 7772 + 2 7 i 2 = 9 
x2 — 4 y = 20 
x2 + y2 = 25 

x 2 + xy = - 6 
xy — y'^ = - 35 
2 x 2 — xy = 28 
x 2 4- 2 y 2 = 18 


26. 


+ y2 = 10 a2 
xy — 3ar 

{ x — y = —3a 
3x+ly = 4a 


27. 


28. 


3 x 2 + y 2 = 2 52 
x2 — 4 y2 = 5 52 

f ^2x2 4- c2y2 = c2 4- ^2 
I xy = 1 


158 


ALGEBRA 


165. Equivalent systems of equations are two different sys* 
terns which have the same common solutions. 


166. An equation is rational if the variables do not appear 
under a radical sign or with an exponent like |. (See § 111.) 

An equation is integral if the variables do not appear in a 
denominator. 

The degree of a rational integral equation is the sum of the 
exponents of the variables in that term in which said sum is 
largest; thus the degree ol — 2 xy = 0 is 3. 

Rule. Two rational and integral equations having two vari¬ 
ables, whose degrees are m and n respectively, have mn com¬ 
mon solutions provided the system cannot be reduced to an 
equivalent system consisting of equations of lower degree. 

Thus, a system consisting of a cubic equation and a quadratic 
equation should have 3 • 2 or 6 common solutions. Some of these 
may be imaginary, and some may be repeated solutions. 


167. Solution of systems which are reducible by division to 

equivalent systems whose equations are of lower degree. 

( 1 ) 

20 ( 2 ) 


Example. 


Q 1 fU f / - 2/4 = 240 

Solve the system < „ , „ 

[x^ y^ = 


Solution. 1. By the rule of § 166, it would appear that there should 
be 4 • 2 or 8 common solutions. 

In this example, however, can be divided by + y"^. 

2. Dividing (1) by (2), — y^ = 12. (3) 

3. Now form the system 


This system will have 2 • 2 or 4 common solutions. 

4. (3) + (2): 2 = 32; = 16; a: = ± 4. 

5. When x = + 4, 16 + z/^ = 20; 2 /^ = 4; y = ±2. 

X = 4:, y = 2, and x = 4,y = —2 are two solutions. 

6. When a: = — 4, 16 + 2 /^ = 20; 2 /^ = 4; 2 / = ± 2. 

X = — 4, y — 2, and x = — 4, y = — 2 are two more solutions. 
Check all four solutions by substituting in equations (1) and (2). 


SYSTEMS INVOLVING QUADRATICS 


159 


EXERCISE 93 

Solve the following systems; check the solutions by substitut¬ 
ing in both equations: 


1 . 

2 . 

3 . 

4 . 

5 . 

6 . 

7 . 

8 . 

9 . 

10 . 


I = 14 

\x + y = 7 
{ — d? — bQ 

\ c — d = 4: 

\ m 

1 x* + 1/2 = 13 ■ 

I = 240 

— v} = 12 
\ x(x — y) = 50 
1 y(^ - y) = 25 
j — S a^b = 54 
\ a — 3 6 = 6 
( x"^ 2 xy — 3 y^ == 17 

\x + 3y = 17 
J - # = 133 
\ c — d = 7 

/2x2+ lla:y + 12y2 = 63 

\2x + 3y = 9 

I = 35 

\ a: + y = 5 


15 . 

16 . 

17 . 

18 . 

19 . 

20 . 

21 . 

22 . 

23 . 

24 . 


f a:^ — = 218 

[ a:^ -f- a:y + = 109 

f + y3 = - 217 
[ a; + y = - 7 
f x^ — x'^y = 54 
I a: - y = 6 


25 . 

26 . 

27 . 


x3 - 8 y3 = 208 
X — 2 y = 4 


28 . 


fa2-62 = 4a + 66-8 

[ a — h = 2 
[ 7n? — v? = — 117 
— n = — 3 
3 a T 5 = 2 

27 + 63 = 9g 

a:^ + xy — 6 y2 = 21 
a;y + 3 y2 = 84 
' X - y = 3 
x^y — xy^^ = 30 
-f 63 ^ 224 
a + 6 = 4 
x^ + xy^ = 84 
X^y ^4 ^ 28 

2 X — y = 3 
2 x^y — xy^ = 15 
p{l + .03 r) = 420 
p{l + .07 r) = 480 

a:3 — y3 = 19 

a:^ + a:y + y2 = 19 

a.3 q, y3 _ ^2 

a^2 — a:y + y2 = 12 

8 - y3 = 133 m3 

2 X — y = 7 m 
2r+ 3s = - 6 

2 4- rs — 3 = — 42 

27 x^ - 8 y^ = 208 f 

3 X — 2 y = 4:t 


160 


ALGEBRA 


EXERCISE 94 

1. Find two numbers whose sum is — 9 and the sum of 
whose squares is 101. 

2. The sum of the squares of two numbers is 1274 and the 
larger is 5 times the smaller. Find the numbers. 

3. The product of two numbers is 5 and the sum of their 
squares is 26. Find the numbers. 

4. Find two numbers whose sum is 10 and whose product is 9. 

6. Find two numbers whose difference is 5 and whose 

product is 6. 

6. The product of two numbers is 48 and the sum of their 
squares is 100. Find the numbers. 

7. The sum of two numbers is 8; the square of the first 
number increased by the product of the two numbers is equ^l 
to 49 diminished by the square of the second number. Find 
the numbers. 

8. The square of the length of the diagonal of a rectangle 
is 40 and the area of the rectangle is 12 square feet. Find the 
dimensions of the rectangle. 

9. Find two numbers such that the square of the first 
diminished by the product of the two numbers shall be 15, 
and such that the difference between their squares shall be 21. 

10. Find two numbers such that their product increased by 
the first number shall be 8, and such that their product in¬ 
creased by twice the second number shall be 12. 

11. The area of a certain triangle is 108 square feet. Its base 
exceeds its altitude by 6 feet. What are the base and altitude 
of the triangle? Also, solve this problem graphically. 

12. If the larger of certain two numbers be multiplied by the 
difference of the two numbers, the result is 44. The product 
of the two numbers is 77. What are the numbers? 


SYSTEMS INVOLVING QUADRATICS 


161 


13 . In 6 hours, two boys row 16 miles downstream and back 
again. Their rate upstream is twice the rate of the current. 
Find the rate at which they row in still water and the rate of 
the current. 

14 . The sum of the squares of the two digits of a number is 
74. If 18 be added to the number, the digits of the sum are the 
digits of the original number in reverse order. Find the number. 

16 . Find the sides of a rectangle if the area of the rectangle 
increased by the shorter side is 15 and if the area increased by 
the longer side is 16. 

16. The sum of the volumes of two cubes is 72; and an edge 
of one cube plus an edge of the other cube is 6. Find the length 
of the edges of each cube. 

17 . Find the number of two digits such that if the digits be 
reversed, the new number minus the original number is 18, and 
their product is 403. 

18 . Find two numbers the sum of whose squares increased by 
the product of the numbers is 28, and the difference of whose 
cubes is 56. 

19 . The area of a trapezoid is 36 square inches. Its altitude 
is 3 inches. Find the two bases of the trapezoid if the square of 
the length of the lower base is 192 more than the square of the 
length of the upper base. 

20. The sum of a certain fraction and its reciprocal is 

If three times the numerator is decreased by twice the denomi¬ 
nator, the result is equal to — 12. Find the fraction. 

21. In an isosceles triangle, the vertex angle is 60° more than 
the square of either base angle. Find the angles of the triangle. 

22. The hypotenuse of a right triangle is 13 inches. Its alti¬ 
tude exceeds its base by 7 inches. What are the base and alti¬ 
tude? Solve this problem also graphically. 


162 


ALGEBRA 


EXERCISE 95. CHAPTER MASTERY-TEST 


1. Determine the common solutions of the system 

{ -{• xy y‘^ = ^ Check the solutions. 

\x-\- y =2. 

2 . Determine the common solutions of the system 

x^ — y‘^ = 12 Properly pair off the values 

^2 _j_ 2/2 = 24. of X and y. 


3. Determine the common solutions of 

4. Determine the common solutions of 
2 x^ — ?> xy = — 4 


2/2 =_ 3 


xy = - 2. 


4 0^2/ — 5 2/^ = 3. 

6. (a) Sketch the graphs for the system 


6 . (a) Sketch the graphs for the system 
(6) How many real common solutions a] 

7. (a) Sketch the graphs for the system 


+ 2/2 = 36 
xy = 4. 

(6) How many real common solutions are there? 

0:2 + 2/2 = 10 
X + 2 / = 3. 

ih) How many real common solutions are there? 

0^2 + 2/2 = 16 
X 2 / = 6 . 

(6) How many and what kind of common solutions are there? 

(c) Verify this statement by solving the system algebraically. 

(d) Check the solutions by substituting in both equations. 


8. The area of a rectangular field is 216 square rods and 
its perimeter is 60 rods. What is its length and width? 

9. The difference of the rates of a passenger and a freight 
train is 10 miles an hour. The passenger train requires 1 hour 
more for a trip of 175 miles than the freight train requires for 
a trip of 100 miles. Find the rate of each. 

10. How many and what possible kinds of solutions does a 
system of two quadratics have? Illustrate by free hand graphs. 


SYSTEMS INVOLVING QUADRATICS 


163 


EXERCISE 96. CUMULATIVE REVIEW 


1 . Find the prime factors of each of the following: 

(а) — 9 + 6 m — 1 (c) (m — nY — 5(m — n) — ^ 

(б) a® - 7 - 8 (d) _ 125 


2. Simplify: 


L _ ah \ 

a — b c\ — ab h^) 


3. Rationalize the denominator and simplify 


2V6 - 3 
V3 - V 2 


4. Draw the graph of the function + 3 x — 2 and from it 
determine the roots ofa:2 + 3a: — 2 = 0 correct to tenths. 

5. (a) Solve the formula s = at + j gf for t. 

(b) Using your results as new formulas, find t correct to 
hundredths when g = 32, a = 1000, and s — 7500. 

6. A picture 15 inches wide and 20 inches long is to be sur¬ 
rounded by a frame whose area shall be f that of the picture. 
What must be the width of the frame? 

7. Solve the equation ax^ 2 bx c = 0 by completing 
the square. 

8. By the formula 8 = ^|2a + {n - l)dj^ determine n 

when S = 25, a = 7, and d = — 1. ^ 

9. A crew can row 12 miles downstream and back again in 5j 
hours. If the rate of the stream is 3 miles per hour, find the rate 
of the crew in still water. 

10. Solve the following system for x and y and correctly group 
your answers; also check the solution: 

r - 7/2 = 16 
I ?/2 — 14 = a: 

11. (g) Determine graphically the common solutions of 
xy = S and x — y = 5. 

(b) Solve the system algebraically. 





XL FACTORS AND EQUATIONS OF 
HIGHER DEGREE 

The Factor Theorem 

168. The factor theorem aids in solving special equations 
of higher degree, such as — 2 = 0. 

Exam-pie 1. Is x — 1 a factor of — 2? If it is, it is 

an exact divisor of cr® + a:^ — 2. 

The division at the right shows that 
cc — 1 a factor. The factor theorem 
enables us to determine this without 
dividing. Take the 1 of x — and 
substitute it in — 2. The result 

is 0. To one who knows the factor 
theorem, this is enough to tell that x — 1 
is a factor of x^ + x^ — 2. 

Statement of the theorem. If the value of a rational and 
integral polynomial in x is 0 when a is substituted for x, then 
X — a is a factor of the polynomial. 

Proof: 1. Let Pix), [read the polynomial in a:] be the divi¬ 
dend; let x — a be a divisor of it; Q the quotient; and R the 
remainder. 

2. Then P{x) = {x — a) ■ Q + R. [See § 19] 

3. Let X = a on both sides. Then P{a) = (a — a) • Q R, 
or P{a) = R. 

Here P(a) means the value of the polynomial when x is a; 
and by the theorem this is to be zero. 

4. /. 0 = R. 

5. Hence x — a is a factor. 

164 


+ 2 X 4-2 


X® 4- 

x® — 


-h Ox - 2 


4- 2 x2 4- 0 X 
2 x^ — 2 X 


2x - 2 
2x - 2 








FACTORS AND EQUATIONS OF HIGHER DEGREE 165 


Example 2. Factor + x^ — 4 .r — 4. 

Solution. 1. Any factor must end in a divisor of 4. Is a: — 1 a 
factor? Does U + P — 4 — 4 = 0? 

No, therefore x — 1 is not a factor, 

2. Is X + la factor? Aozi? we must substitute — 1 since x — (— 1) 
= X + 1. Does — 1 + 1+ 4 — 4 = 0? Yes. 

Therefore x + 1 is a factor. 


3. The remaining factor is found by 
division at the right. 

4. x^ + x^ — 4 X — 4 
= (x — l)(x2 — 4) 

= (x - l)(x + 2)(x - 2) 


x^ - 4 _ 

x + 1 |x^ + x2 — 4x — 4 
x^ + x2 

- 4x - 4 

— 4 X — 4 


Note. On the next two pages is a clever short method of division 
which may he studied before doing the next exercise, or which may he 
omitted entirely. 


EXERCISE 97 

Find the factors of: 


1. 


- 2 

X — 4 

11. 

X3 - 

■ 4 x2 + X + 6 

2. 


+ X 

- 2 

12. 

2/3 + 2 7/2- 

- 9 7/ - 18 

3. 

X? 

+ X 

+ 2 

13. 

mf - 

- 5 m2 

+ 3 m + 9 

4. 

x^ 

- 2 

x2 - 9 

14. 

7-3 + 

■ 3 r2 - 

- 4r - 12 

6. 

x^ 

+ 2 

X + 12 

15. 

^3 — 

5^2 _ 

- 2 5 + 24 

6. 

x^ 

— X 

- 6 

16. 

2x3 

+ 6 x^ 

' + 3x - 2 

7. 

3x3 + 

4x - 7 

17. 

w? - 

- m2 - 

- 3 m + 2 

8. 

2x3 + 

3x + 5 

18. 

X3 - 

- 19 X 

- 30 

9. 

2x3 + 

X — 3 

19. 

yS _ 

■ y‘^ - 

5 7/ - 3 

10. 

2x3 - 

5 X — 6 

20 . 

2s3 

-3^2 

- 3z+ 2 


169 . In Step 3 of the proof on page 164 there is proof of 
another useful theorem. 

The remainder theorem. The value of a rational and integral 
polynomial in x when x equals a is the remainder obtained when 
the polynomial is divided by x — a. 






166 


ALGEBRA 


170. Synthetic division is a short form of division which 
can he used when the divisor is a binomial. It shortens the 
computation necessary when applying the factor theorem. It 
can be studied in advance of Exercise 97, or can be omitted 
altogether. 

Example 1. Divide 5 ~ 14 — 10 by a: — 3. 


Ordinary Method 
5 re + 3 

re — 3 1 5 rc^ — 14 re^ + 0 re — 10 
5 re^ — 15 re^ 

re2 + 0 re 
re^ — 3 re 

Sl- 10 
3re - 9 

Remainder: — 1 


Synthetic Method 


X -|- 3 


5 — 14 + 0 X — 10 

+ 15 +3+9 

5 +1 +311-1 


Quotient: 5 x^ + x + 3 
Remainder: — 1 


Explanation of the Synthetic Method: 

1. The divisor x — 3 is changed to x + 3. (This permits addition 
in Steps 3 to 5, instead of subtraction as in the usual method.) 

2. 5 x^ ^ X = 5 x^. Write 5 in the third line. 

3. (+ 5) • (+ 3) = + 15. Write it below — 14; add, getting + 1. 

4. (+ 1) • (+ 3) = + 3. Write it below 0; add, getting + 3. 

5. (+ 3) • (+ 3) = + 9. Write it below — 10; add, getting — 1. 

5, + 1, and + 3 are the coefficients of the quotient. 

Hence, the quotient = 5 x^ + x + 3. Remainder = — 1. 

These results agree with the ones secured by the ordinary method. 


Example 2. Divide 7 x^ — 29 — 3 by x + 2. 


Solution. Change x + 2 to x — 2. 


X — 2 


7 x4 + 0 x^ - 29 x2 + Ox - 3 
- 14 +28 +2-4 

7-14 - 1 +2II-7 


Quotient: 7 x® — 14 x^ — x + 2 
Remainder: — 7 


Check. 

7 x^ — 14 x^ — X + 2 

__X + 2 

7 x^ — 14 x^ — x^ + 2 X 

+ 14x3 - 28x^ - 2x + 4 

7 x^ - 29 x2 +4 

- 7 

7 x4 - 29 x2 - 3 

















FACTORS AND EQUATIONS OF HIGHER DEGREE 167 

EXERCISE 98 

Divide by synthetic division, and check: 

1. x^ + 2x‘^-2x-{-5 hyx-1 

2. - 4: y -h Q by y + 1 

3. x^ + 2 x^ — X — 2 by X — 2 

4:. x^ — 3 x^ — X S by a: — 3 

5. 2x^-Sx^+x-8 hyx-2 

6. — 6 by a: + 2 

7. 3 - 8 f - 27 hyt-S 

8. a:^ + 2 a:^ — 5 a:^ — 12 by a: + 2 

9. 5 + 6 7-2 - r - 12 by r - 2 

10 . 3 a:^ + 11 a:^ — 4 a:2 — a: — 8 by a: + 4 

11. Is a: — 1 a factor of a:^ + 3 a:^ — 4? 

Suggestion. Divide. What should the remainder be ? 

12. Is a: + 2 a factor of a:^ + a:^ + 4? 

13. By § 169, if + a:2 — a: — 2 be divided by a: — 1 the 

remainder is the value of a:^ + — a: — 2, when 1 is substituted 

for X. Find this value by finding the remainder by synthetic 
division. Check it by substituting 1 for x in — x — 2. 

As in Example 13, find: 

14. The value of x^ + x^ — x — 2 when x = + 2. (This 
means, divide by x — 2, and find the remainder.) 

16. (a) The value of 2 x^ — x^ + x — 4 when x = — 1. 

(6) Also when x = + 1, (c) when x = + 2. 

16. (a) The value of x^ — x — 6 when x = 2. 

(b) Tell two factors of x^ — x — 6. 

17. (a) Divide x^ — 1 by x — 1. 

(6) Is X — 1 a factor of x^ — 1 ? 

18. (a) Is X — 1 a factor of x^ — 1? 

(6) Is X + 1 a factor of x^ — 1? 


168 


ALGEBRA 


171. Graph of a third degree function of one variable. 

Example. Draw the graph of — 4 — 2 a: + 8. 


Solution. 1. Let y^x^ — 4:X^ — 2x-\-8 

We must find values of y when a; = 0, 1, 2, 3, 4, etc. 

This can be done by direct substitution or as in § 170 as a con¬ 
sequence of the Remainder Theorem, § 169. 


For example: when x = 4 


1 - 4 - 2 + 8 

+ 4 0-8 

1 0 - 2 II 0 


y = Q when x = 4 


When X = 

0 

1 

2 

3 

4 

5 

- 1 

- 2 

then y = 

3 

8 

- 4 

- 7 

0 

23 

5 

- 12 



The graph of every cubic function of one variable has this same 
general shape. 

2 / = 0, where the curve crosses the x-axis. Hence 

X = — 1.42 (at A); x = + 1.42 (at B); x = + 4 (at C) are the 
roots of the equation x^ — 4x2 — 2x + 8 = 0. 

The line y = — 5 cuts the curve at D, and F. 

At D, X = — 1.65; at x = 2.2; Sit F,x = 3.5. Therefore — 1.65, 
+ 2.2, and -}- 3.5 are the roots of the equation x^ — 4 x^ — 2x4-8 
= — 5. The line 2 / = — 8 cuts the curve only at G, where x = — 1.8. 

Hence x = — 1.8 is the only real root of x^ — 4 x^ — 2 x 4- 8 = — 8. 

The other two roots are imaginary. 



























































































































FACTORS AND EQUATIONS OF HIGHER DEGREE 169 


172. Algebraic solution of equations of higher degree is 
difficult except for special equations. 

It is proved in higher mathematics that every equation of 
degree n has n roots. 

Example 1. Solve the equation + 4 — 5 = 0 

Solution. 1. Factoring: {x^ — l)(a:2 + 5) = 0 
2. .’. — 1 = 0; = 1; a: = rb 1 

or + 5 = 0; = — 5; a: = + 5; x = — V— 5 

The four roots are: +1; —1; ± V^S; ±i^5. 


Example 2. Solve the equation + a: — 2 = 0 


Solution. 

2 . 

3. 


1. By the factor theorem (a: — l)(a:2 + x + 2) = 0. 
aj — 1 = 0; or a;2 + a; + 2 = 0 

- 1 ± Vl - 8 

. . a: = 1; or a; = -jr- 


4. Hence x = 1; x = ——x = 

EXERCISE 99 

Solve the following equations: 


1. 

x^ — 

5 + 4 = 0 



14. 

X3 

4- 

2x2 

— X — 

2 = 0 

2. 

— 

26 c2 + 25 = 

0 


16. 

X^ 

- 

2 x2 

4- 1 = 

0 

3. 

y4 _ 

29 2/2 

8 

+ 


0 

16. 

a4 

- 

2 X 

4- 1 = 

0 

4. 

x4 - 

8 x2 - 

-9 = 0 



17. 

x^ 

- 

3 x2 

4- 2 = 

0 

6. 

y4 _ 

- 

- 36 = 

0 


18. 


4- 

3 2/2 

- 2 = 

0 

6. 

t^ + 

2f2 - 

8 = 0 



19. 

f 

- 

4 2/2 

- 72/ 

4- 10 = 0 

7. 

4 x"^ 

— 4 x2 

4- 1 = 

0 


20. 

f 

4- 

3 2/2 

- 4 = 

0 

8. 

2y^ 

-9 2/2 

-f 4 = 

0 


21. 

23? ■ 

■f X2 

— 6 X 

+ 3 = 0 

9. 

4 m^ 

- 17 

^2 -h 4 

= 

0 

22. 


- 

^2 - 

8z4- 

8 = 0 

10. 

6 

— 5 x2 

+ 1 = 

0 


23. 

x^ 

- 

a4 - 

■ 8 x^ + 12 X = 0 

11. 

x^ — 

x2 — i 

5x4-5 

= 

0 

24. 

C3 

- 

6c2- 

■f 11 c 

-6 = 0 

12. 

x^ — 

2x2 - 

- X 4- 2 

= 

0 

26. 

a4 

4- 

X2 - 

■ 6 X = 

0 

13. 

yZ _ 

2 2/2- 

- 62 / - 

3 

= 0 

26. 

23? ■ 

- 5a 

c2 4- X • 

+ 2 = 0 






XII. EXPONENTS AND RADICALS 


173. We shall first prove the five laws of exponents for posi¬ 
tive integral exponents. So far you have used these laws without 
having proved them. 


174. Definition. When m is a positive integer, 

x'^ = X ' X ' X . X. {m factors) 

[Notice how this is written. The dots are read ‘^and so on.’" 
The number of factors is indicated at the right or below.] 


175. The multiplication law. x"* • x” = x"*+” 

Proof. 1. x^ = X ' X ' X . X (m factors) 

2. x^ = X - X - X X {n factors) 

3. • ic” = [x ' X ' X . x] ’ [x ' X ' X . x] 

(m factors) {n factors) 

4. = X ' X ' X . X ’ X ' X ' X . X {m n factors) 

5 x^ * x” = x^^^ 

176. The division law. x"* -j- x” = x"*~” (m greater than n) 

Pj.Qof 1 ^ ^ . ^ ^ ^ .^ factors) 


2 . 


3. 


•/. / 

= X • X . X. 


{n factors) 
(m — n factors) 


177. The power of a power law. (x"*)” = x"*” 

Proof. 1. (x^)^ = x"^ ’ x^ ' x^ ' x^ x”* (n factors) 

2 . (x”*)" = .+«» (n addends in exponent) 

3. (x”^)^ = x^^ 

170 





EXPONENTS AND RADICALS 


171 


178. Power of a product law. {xy)^ = x^y^. 

Proof. 1. {xyY = {xy){xy){xy) . {xy) (n factors) 

2. /. (xyY = {x • X ‘ X . x) (y • y ’ y . y) § 13, b 

(n factors) {n factors) 

3 . {xyY = x^y^ 

/ x \^ 

179. Power of a quotient law. = — 

2 • ^ ^ ^.^ factors) 

W y ■ y ’ y . y {n factors) 

3. (-Y = — 

\y/ 2 /” 

EXERCISE 100 

In the following examples, the literal exponents represent 
positive integers. Find the results of the indicated operations: 


1. 



12. 

2^2 - 

2” 

23. 

(2^)3 

34. 

( — 4 a^by 

2. 

m® • 


13. 

-f- 


24. 

(w^y 

36. 

(- i af 

3. 

2/4 . 

2/^ 

14. 

QtSift ^ 

r a::”" 

26. 

(x^y 

36. 

{~%cdj 

4. 

iC® • 

a:^ 

16. 

2^777+4 

^ y^ 

26. 

{fy 

37. 

(f xy‘‘y 

6. 

^2n . 

^5n 

16. 

2m ^ 

2m-l 

27. 

{m^y 

38. 

{x'^y’^w) ^ 

6. 

X^ • 

X^ 

17. 

- 

^ A:*- 

28. 

(- a?hy 

39. 

(— a?h^y 

7. 

2/C-2 

.2/3 

18. 

^3r+l 

-J- 

29. 

(- 2 a6)3 

40. 

{a^y 

8. 

^2n . 


19. 

^71+3 , 

-J- a:2 

30. 

(2 xyY 

41. 

(b^y 

9. 



20. 

a,n-3 

x^ 

31. 

(3 xy'^zy 

42. 

(xYY 

10. 


. ^m+1 

21. 

(x^y 


32. 

(4 mn2)3 

43. 

{r^s^Y 

11. 

2/10. 

- 2/^ 

22. 

(fy 


33. 

(—2 x‘^yy 

44. 

(x^yny 

46. 

© 

) 

46. 

&: 

i 

47. 

m 

48. 

/ 2 m^Y 

V ■^/ 




172 


ALGEBRA 


180. Roots of numbers: 

(а) You know that a number like 9 has two square roots; 
that + 3 is its principal square root; and that the principal 
square root of 9 is indicated by V9 or by 9^. (See p. 100.) 

(б) Similarly a number has three cube roots, four fourth roots, 
and n nth roots. 

Thus, + 2 is a cube root of + 8, since (+ 2)^ = + 8. 

The other two cube roots of + 8 are complex numbers. + 2 is the 
principal cube root of + 8. 

— 2 is a cube root of — 8; since { — 2Y = — 8. 

The other two cube roots of — 8 are complex numbers. 

~ 2 is the principal cube root of — 8. 

Definition. The principal nth root of a number x is indicated 
by '^x. n is the index of the root; V is the radical sign; x is 
the radicand. 

The principal root of a positive number is positive. 

Thus = + 5; ^ = + 2. 

The principal odd root of a negative number is negative. 

Thus = - 2; </- 32 = - 2; 27 = - 3 x\ 

181. Roots of numbers are also indicated by fractional ex¬ 
ponents. 

(а) Why the exponent J is used to indicate the square root. 

If xh obeys the multiplication law of exponents, then (x^) (a:l) 
should = x^, or x. Then xh ought to mean ^/x, or — 

We define xi as being the same as V^. 

(б) Similarly, if (a:i) (a:i) (xl) = x^, or X, then x\ ought to be a 
cube root of x. 

We define x\ as being the principal cube root of x. ('V^). 

(c) In general, we define xn as being x'^. 

Thus 4f = V 43 = = 8; 8l = ^ = 4. 




EXPONENTS AND RADICALS 


173 


EXERCISE 101 

Give the principal root indicated below: 


1. 

27l 

7. 

YIe 

13. 

(— a^x^)i 

19. 

(tb" 

2. 

16^ 

8. 

(144)i 

14. 


20. 

Yii2 

3. 

64i 

9. 

81} 

15. 

{xY)^ 

21. 

(64)1 

4. 

i 

10. 

(- 125)1 

16. 


22. 

Y.25 

6. 

lOOl 

11. 


17. 

(i a®)l 

23. 

(.36)1 

6. 

(- 27)1 

12. 

Him 

CM 

CO 

18. 

Yi6 a* 

24. 

(.008)1 


■m' 




16\i 


32 


a 

/- 32\i V 

^27 (^) “ “ Y 


Yf 

(f)' 


34. 


35. 


(t)‘ 

- 27 


182. By definition x” means 
also means 

For example: to find 4^, find (Vi)3, or 2^, or 8. 


It can be proved that it 


EXERCISE 102 

Simplify: 

1 . 9t 4. 27f 7. (- 8)1 10 . (- 32)t 

2. 16f 6. 4i 8. (- 27)t 11. (- 8)1 

3. 8t 6. 83 9. (- 32)1 12. (- 27)1 

183. (Optional.) What are the cube roots of 1? 

1. Let X = or = 1; or x^ — 1 = 0 

2. (x — l)(x^ + X + 1) = 0 

3. Completing the solution of this equation: 

xi = 1; X2 = i (“ 1 + ^V^); X3 = 1 ~ iV3). 

Note. The interesting fact is that the cube roots of 8 are: 

2; 2 [K- 1 + ^’^3)]; 2 [§(- 1 - fv's)]. 







174 


ALGEBRA 


184. What meaning has the exponent zero? 

If obeys the multiplication law of exponents then 
or x^. x^ = x^ ^ x^, or x^ = 1 . 
This suggests the following definition. 

Definition. x® = 1, except when x = 0. 

Thus 50 = 1; 50000 = 1; (- 15)o = 1. 

185. What meaning has a negative exponent? 

If x ~~^ obeys the multiplication law of exponents, then 

^-3 . ^+3 = = x ^ = 1. Hence x ~^ = 

This suggests the following definition. 

Definition. x“” = — 

Xn 


Thus: 



1 

8 ' 


4-t = _ 

41 


1 

2 


EXERCISE 103 

Find the numerical value of, or the simplest form of: 


1. 

40 

11. 

3-® 

21. 

T 

GO 

31. 

18" • 2-< 

2. 

yO 

12. 

2-5 

22. 

16-^ 

32. 

3-3 • 27 

3. 

3x0 

13. 

6-2 

23. 

27-i 

33. 

1000" • 2-3 

4. 

(2 x)o 

14. 

(- 3)-i 

24. 

(- 8)-i 

34. 

10-3 

5. 

20 X 

15. 

(- 3)-2 

26. 

25~l 

36. 

4-# 

6. 

4-2 

16. 

(- 2)-5 

26. 

0 

0 

CO 

36. 

32 X 10-' 

7. 

3-1 

17. 

(+ 2)-^ 

27. 

64 • 2-3 

37. 

32 X 10-3 

8. 

3-2 

18. 

(- 3)« 

28. 

9» • 9-i 

38. 

56 X 10-3 

9. 

5-2 

19. 

(- 2)-^ 

29. 

(64)-i 

39. 

5 X 10-3 

10. 

80 

20. 

(- 4)-^ 

30. 

(- 27)-i 

40. 

1 X 10-3 

Rewrite using positive exponents 




41. 


42. 

2r-< 

43. 

2 arV 

44. 

0 

01 


EXPONENTS AND RADICALS 


175 


186. Negative exponents in fractions. 


Example 1. 


3a-^ 

4 c“* a® ’ 4 4 


Observe that a~^ of the numerator becomes in the de¬ 
nominator; that c~^ of the denominator becomes in the 
numerator. 

Rule. Any factor of one term of a fraction may be trans¬ 
ferred to the other term provided the sign of its exponent is 
changed. 

Example 2. 5^^ ^ 5^. ■ 

w~H yH 

Example 3. ^ = 3 

cd^ 

EXERCISE 104 


Rewrite with only positive exponents : 


x-^y 

5 a:-3 

4. -pr- 

5 m^n~^ 

10. 

k 


y~^ 

6r2 


10-15 

3 m 

mn 


11. 

10-V 

n~^ 

' 10-^ 




4 a~^ 

6 

- 8oV* 

12. 

4 • 3-'7 

63 

■ (2 3/)-^ 

* 1 

r“3 


Write without any denominator: 


13 

_ 2 mV 

^5 

17 Z_^ 

m/ 

19. 

5 a^h~^ 
2-%-2 

CO 1 

16. 


20. 

6 a* 

d-^ 

z^ 



r* 


21 . 


1 

(1 + iY 



i 


24 . 


1.02^ 


^ - 1 -1- 1 -1- 1 - \- -i— 

1 . 02 ^ 1.023 ^ 1.022 ^ 1.02 


[ Examples 21-24 
from the mathematics 
investment. 


: are~] 
ics of 





















176 


ALGEBRA 


187. Meanings have been given to fractional, negative, and 
zero exponents so that each will obey the multiplication law of 
exponents. It can he proved that, with these meanings, they 
obey all the laws of exponents given in § 175 to § 179, on pages 
170 to 171. • The following exercise gives a little practice in 
use of these laws with these new exponents. Examples 30 to 50 
are preparation for the study of the next chapter. 





EXERCISE 105 



Perform the indicated 

operation: 



1. 

^-3 . ^4 

11. 



21. 

(x3)-2 

2. 

S~‘^ • 

12. 


- 

22. 

(Xi)2 

3. 

x~^ • xy'^ 

13. 

x~^ - 


23. 

(X“l)2 

4. 

1 1 

X2 • XZ 

14. 

x~^ - 

^ x^ 

24. 


6. 

1 3 

16. 

X -r- 

1 

X2 

26. 

(x%i)2 

6. 

2 1 

XZ • X^ 

16. 

x^ 

1 

X^ 

26. 

{x-hyhy 

7. 

m~h • m 

17. 

X2 ^ 

■ ^3 

27. 

{xl • y-hy 

8. 

n~^ • ni 

18. 

2 . 

yz -v 

• yi 

28. 

{x~i)~i 

9. 

n~i ■ n~i 

19. 

x~l 

. _i 

-h X 4 

29. 

{x-jy-iy 

10. 

x~^ ■ x~^ 

20. 

x~i 

. _i 

-j- X 3 

30. 

(x-f) 1 

31. 

1(F X 102 

36. 

102.6 

X 101-2 

41. 

102:63 ^ iQi 

32. 

10* H- 10* 

37. 

10.25 

X 10-22 

42. 

10.75 ^ 10.25 

33. 

102 ^ 103 

38. 

101.250 X 101 

43. 

101.36 ^ 10.45 

34. 

10-2 X 10-1 

39. 

102.73 

1 X 101-24 

44. 

10-7 10-8 

36. 

10-2 ^ 10-2 

40. 

10301 

X 10-477 

46. 

109.27 X 10-84 

46. 

(102)i 

61. 

2x . 

2v 

66. 

3x . 33x ^ 32 x 

47. 

(52-*)^ 

62. 

22a . 

2“ 

67. 

101-25 X 100 

48. 

(102-25)i 

63. 

4c ^ 

4d 

68. 

102-1 ^ 100 

49. 

(101-462)4 

64. 

3” 

3^ 

69. 

101.50 ^ 100 

60. 

(109.762)i 

66. 

5^ 

52 x • 

60. 

102-75 X 1000 


EXPONENTS AND RADICALS 


177 


Operations with Radicals {Optional) 

188. When an indicated root can be obtained, the result is a 
rational number; when the root cannot be obtained, the result 
is an irrational number, or radical. 

The denominator of the fractional exponent, or the small 
number written in the angle of the radical sign, is the index of 
the radical. 

Thus the index of V5 is 2; of '^4 is 3; of 2i is 3. 


189. 

I. 

II. 

III. 


To manipulate radicals easily, remember: 

The definition: {^G:)^ = x. Thus {'^bY = 5. 
The principle: • ^6; or {ah)n = 


The principle-. 





an 

b- 


an 




IV. That radicals can be indicated by fractional exponents, 
which obey all the laws of exponents. 


190. Simplifying a radical: 

(a) A radical is usually written with as low an index as possible. 
Thus: = (82)i = Si = VK 


(6) Factors are removed from the radicand when possible. 
Thus: , = V25 • 3 = . V3 = 5V3 

</72 ‘ ■ 9 • v^2 = 2 

(c) The radicand is written without a denominator. 

Thus: 

3/~3~ ^ ^ 3 ■2a ^ ^ 

^ 4 T 8 flS 2 a 


5 ^ J 5 • 3a ^ Vl^ ^ Vl5 a 
12 a ^ 12 a ■ 3 a a/qa ^2 6 a 


Observe: If the radical is a square root, the denominator is 
made a perfect square; if it is a cube root, the denominator is 
made a perfect cube; etc. 












178 


ALGEBRA 


191. Rule. To reduce a radical to its simplest form: 

(а) Change it to an equal radical with lowest possible index. 

(б) Remove from under the radical any perfect power factors. 
(c) Change the radical to one in which the radicand is not 

fractional, as in ^ 190, c. 

EXERCISE 106, a 
Simplify the following radicals: 


1. 


3. 


6. 


7. 

^2511 

2. 


4. 


6. 


8. 

^9 0^62 

9. 

vis 

11. 


13. 

-^72 

16. 

V28 a^b 

10. 


12. 


14. 

-^32 

16. 

-^24 a4 

17. 

/I 

19. 

1/- 

y 8b 

21. 


23. 

'l/— 

V 18 

18. 

\/~i~ 

20. 

n 

22. 


24. 


26. 


27. 

^16 <« 

29. 


31. 


26. 


28. 

-^27 

30. 

V 98 rs^ 

32. 

V32 m* 

33. 

/i 

36. 

V 5x 

37. 

y? 

39. 

3/4^ 

V 9t 

34. 


36. 

V 4b 

38. 

t/s, 

40. 

/c-d 

V 0+ d 




EXERCISE : 

L06, b 



1. 


6. 

V225 

11. 

■^128 

16. 


2. 


7. 


12. 

^108 a’’ 

17. 

A^16 x^m^ 

3. 


8. 


13. 


18. 

-^80 

4. 

•'5^32 xY 

9. 

"Mii 

14. 

aV . 5 ^4 

^56 

19. 

^128 

6. 

"V^64 x^y 

10. 


16. 

-^243 a:" 

20. 

'^xy'^z 

21. 

i/x 

y ^x^ 

22. 

(/? 

23. 

1/— 

K 16 6^ 

24. 

6 / a22^7 

K 32 a:^ 


























EXPONENTS AND RADICALS 


179 


192. Addition and subtraction of radicals. 

The sum of two radicals like V2 and Vs can be indicated 
thus: V 2 + Or, each can be expressed decimally and 

the approximate sum be obtained. 

Thus: \/2 + V 3 = 1.414 + 1.732, or 3.146. 

In order that radicals may be combined, they must be similar 
radicals. Similar radicals are radicals which have the same 
index and which have the same radicand. 

Example 1. + 3V6 = V4~6 + 3 V 6 = 2 V 6 + 3V^ or 5 Ve 

Example 2. - >^27 • 2 = h</2 - or - 


EXERCISE 107 

Simplify and combine similar radicals: 


1. 

V 32 + V 72 

16. 

2. 

2 V18 X — V 5 O X 

17. 

3. 

V128 a ~ a 

18. 

4. 

v'i + iVTs 

19. 

6. 

V 75 m — V 27 w 

20. 

6. 

5 V 1 

21. 

7. 

3 aVWa - V144 a? 

22. 

8. 

2 a:V25 xy^ — 3 yV4 

23. 

9. 

TiV8 — mV32 mn^ 

24. 

10. 

VT6+^ 

25. 

11. 

V 54 a: — V 16 X 

26. 

12. 

- V 241/2 

27. 

13. 

Vf + Vf 

28. 

14. 

vV« - 

29. 

15. 

3xVrx + 5 V 3 I? 

30. 


+ Vie 
V48 - VI 
V 2 + V64 



+ Vx 

- xVTs + Vsox^ 
yVW, - V243 y‘‘ + yVlb 
3VT1 - V44 + V99 
5^4 + 

V 48 - 3V12 + sVi 

- J aVs 
3 V|^ - + 2 Vl 8 a 

Vltx - VIIx + 2 VTx 

2 Vf X — Vl- X — 

— Vr% + rVl6 « 












180 


ALGEBRA 


193. Multiplication of radicals. 

Example 1. 2 V 3 • 3^6 = 6V^ = = 6 • 3^2 = ISV^ 

Example 2. 4v^2 • 5^4 = 20^^ = 20^8 = 20 • 2 = 40. 
Example 3. V 3 * >^3 = 3^ • 3^ = 3^ • 3^ = (3^ • 3^)^ = 


EXERCISE 108, a 
Find the following products: 

1. V3 • Vis 4. 3V6 • 2V2 7. 2V7 • 3\/7 10. (V^)2 

2 . \/2 • Vli 6 . 3V3 • 8 . VT2 • 3V3 11 . ( 2 V ^)2 

3. 2 V 7 • 6. V 5 • a/15 9. 12. (3V^)2 


13. Illustrative Example: (2 + \/3)(3 - 

= 6 + 3V3 - 

14. (2 - V^)(2 + V^) 19. 

16. (7 - V^)(7 4- V^) 20. 

16. (V^ + 3)(vT^ - 3) 21. 

17. ( 2 V 3 - 1)(2a/3 + 1) 22. 

18. (3 + V3)(3 - V 3 ) 23. 

24. Does 1 + V 2 satisfy the equation x‘^ — 2 x — 1 = 0? 
26. Does V 3 — 1 satisfy the equation ^2 + 2 x = 2? 

26. Is 2 — VlO a root of ^2 _ 4 3 . = 0 ? 


V3) 

2V3 - 3, or 3 + V3. 
(2V3 + 1)(V3 - 1) 
(3V2 + 2)(V2 - 3) 
(5 - 2V3)(1 + 2V3) 
(4 - V5)(2 + Vs) 
(V3 + 1)(V3 + 1) 


Find: 


EXERCISE 108, b 


1. {Vx + 1 - 2)2 4. {2-\-Vx- 5)2 

2 . (Vh - 3 + 4)2 5 . (Vi/ - 3 - 2)2 

3. (5 — Va: — 1)2 6 . (Vi — Va: + 1)2 

7. V 9 • Ve 11 . V 3 • V^ 16 . V 2 • Vi 

8. Vie • Vl 2 12 . V6~i • Vs^ 16 . Vi • Vi^ 

9. V3^ . V^ 13. VTi . Vs^ 17. Vk) • V^ 

10 . V6~i^ • V 9 14. Vi • Vie 18 . V 9 • V15 








EXPONENTS AND RADICALS 


181 


194. Division of radicals. 


Example 1. V27 -f- Vq = V'27 9 = Vs. 

Example 2. 3-t-v'2 = 4=X^ = 

\/2 V 2 2 


By multiplying the denominator by itself, we rationalize the de¬ 
nominator. We can now substitute the va!lue of V2 and simplify. 


Example 3. 


V2 

V2 - 1 


V2(V2 + 1) 
(V2 - 1)(V2 + 1) 


2+ V2 
2 - 1 


= 2 + V2. 


In this example, the difference of two numbers in the denomi¬ 
nator was multiplied by the sum; the product is rational. 
(V2 — 1) and (V2 + 1) are called conjugate expressions. 


EXERCISE 109, a 

Find the value of; 


1 . 

2 . 

11 . 

16. 


V 21 

3. 

V 5 O X 

6. 

Vs 

7. 

VsOa* 

9 . 

V14 

VT 

V 2 V 

Vi 

V6a 

V\y 

V^ 

4. 

aV^ 

6. 


8. 

V24ar5 

10 . 

2 VIo 

V -5 

aVs 

Vxy 


VI 

2 


3 


4 


9 


2 X 

Vs 

12. 

Ve 

13. 

V 12 

14. 

Vis 

16. 

V 6 X 


I -V2 Vs+ 2 3 - V5 Ve + 3 


20. Find the value of —— 

X + 1 

(a) when x = Vs - 1; (6) when x = 2 - V2. 


EXERCISE 109, b 


V6 

V2 


Vl2 

Vs 

V2 

6 

Vl 

Vm 

V2 

4- Vio 
VI 

Vl5 

VIo 

10 

10. 

V^ 





















182 


ALGEBRA 


{d) ^-^xhj 


id) \ - 


EXERCISE 110. CHAPTER MASTERY-TEST 

1. Give the numerical value of: 

(a) 100^ (6) (- 27)i (c) (d) 2-^ (e) 16" 

2. Give the simplest form of; . 

(a) (6) </27 xy^ (c) 

3. Give the simplest form of: 

W(/i w/p Wt/ji _ 

4. Simplify: (a) V72 — V32 (b) ^24 x — ^54 x 

5. Simplify: (a) + V3^ a (b) '^^x — ^§x 


fx 

6. Find: (a) V's • Vd (b) 2VYx • 3V14 x 

7. Find: (a) (V2 + x)(V2 - x) (b) (V7 - 3 1 /)(V7 + 2y) 

8. Does 2 — Vs satisfy the equation — 4 a: — 3 = 0? 

eV^ ,,, 9 ,, Vl5 Vis 


8 


9. Find: (a) 


3 V 5 


-6 ^3 


(d). 




2 ^j _ 

10. Find the value of-—— when x = \/2 — 3. 

X + 2 


11. Perform the indicated operations. 

{a) ( 5 - 3 ) (s+2) (c) ( 2 /f)(y-i) (^) 102-^ X 100 

{b) (xi)(xl) {d) (m-i)(m0) (/) (x -2 + 7/-2)(a;-2-f 2/-2) 

12. Perform the indicated operations. 

(а) (r^) -f- (r^) (c) (y+l) ^ (yi) (e) 3^^ H- 3^ 

(б) (x+2) ^ (x“®) {d) (2“l) ( 2 +i) (/) 10^-^® 10 '^^ 


13. Give the simplest form for: 

(a) (102)3 (c) (10-1-3)2 {e) (xf)3 

ib) (101-3)2 (d) (x3)-2 (/) {f)\ 

14. Give the simplest form for: 

(а) 9t (c) 161 {e) (- 8)t 

(б) 8t (d) 25t (/) (- 27)t 


ig) 

(h) {y-h-^)i 


ig) 

ih) (- ix6)4 


4 










EXPONENTS AND RADICALS 


183 


EXERCISE 111. CUMULATIVE REVIEW 


1 . Find the prime factors of: 

(a) (c) ?/' - 6 2/' - 3 2 / + 8 

(fe) + 2 + x^y^ (d) x^ + 32 y^ 

2 . Divide 6 a ^ — a~^ — 27 a~^ + 20 by 3 a~^ — 5. Express 
the quotient with positive powers of a. 

3. Simplify 15^^ • V' 18 x^^ • (3 

4. Rationalize the denominator (or perform the indicated 
division): 


(a) 


4 m 

V2m 


ib) 


3 a 
Wa 


ic) 


Vg 


V3 - V2 


5. Solve for x to the nearest hundredth: x^ — 2.5 x = 1.5. 


6 . Solve the following system; group your results. 

f X - 2/ = 1 
[ x2 + X 2 / + 2 /^ = 37 

7. Solve and check: VOx^+h — 3x=l. 

8 . (a) Without solving the equation determine the nature of 

the roots of 5 x^ — 4 x — 3 = 0. 

{h) What is the sum of the roots? The product? 

9. Determine graphically the roots of the equation x^ — 5 x 
= 2 correct to the nearest tenth. 


10. Determine graphically the common solution of: 

(a) The system { 3 ^ ^ J I 3 (i) The system { = ^ 

11. A flower garden contains 1500 square feet. It is sur¬ 
rounded by a path which is 3 feet wide. The area of the path 
is 516 square feet. What are the dimensions of the flower 
garden? 

12. Ax^ + By'^ = 1 is the equation of a graph which goes 
through_the points x = 4^2; 2/ = 3; and also the point x — 2 , 
y = gV 2. Find A and B. 







Xm. LOGARITHMS 


195. Logarithms are special exponents. 

Every positive number can be expressed exactly or approxi¬ 
mately as a power of 10; as 100 = 10^. 

The exponent required is called the logarithm of the number 
to the base 10. 

Thus 2 is the logarithm of 100 to the base 10. 

It is written briefly thus: 2 = logic 100, or 2 = log 100. 


196. How some logarithms can be obtained. 

10 ° = 1, qr^O = log 1. 10^ = 10, or 1 = log 10. 

10-6 = VlO = 3.162; or .5 0 = lo g 3.162. 

10-25 = (io-5)i = VlO-s = V3.162 = 1.778; or .25 = log 1.778. 
By similar computation, the table below can be obtained. 


197. How logarithms are used. 


Example 1. Find 3.1623 X 17.782. 


Solution. 1. 3.1623 X 17.782 

2. = 10-50 X 101-25 

3. = 101-75 

4. = 56.234 (in the table) 

5. 3.1623 X 17.782 = 56.234. 

The solution is approximately correct. 

Example 2. Find (5.6234)2X 31.623-J-17.782. 

Solution. 1. (5.6234)2 X 31.623 h- 17.782 
2. = (10-75)2 X 101-50 ^ 10125 

3 _ ]^Q1.50+l.B0-1.25 = 101-75 

4. (5.6234)2 X 31.623 ^ 17.782 = 56.234 


10-00 = 1.0000 
10-25 = 1.7782 
10-50 = 3.1623 
10-75 = 5.6234 
101-00 = 10.0000 
101-25 = 17.7820 
101-50 = 31.6230 
101-75 = 56.2340 
10200 = 100.0000 


This solution also may be checked by ordinary computation by one 
who has a lot of time and ambition. 

184 




LOGARITHMS 


185 


198. Logarithms of numbers to the base 10 are called common 
logarithms or simply logarithms. 

199. Graph of the functional relation y = log x. 

Besides the values of the table on page 184, observe: 

10-25 ^ = 10-75 4 - 10^ = 5.6234 ^ 10 = .562. 

Similarly lO-^o = lO'^ 4- 10^ = 3.1623 4 - 10 = .3162. 

Express the values of x correct to tenths, and the values of 
y correct to hundredths. 


Whenx = 

.3 

.6 

1.0 

1.8 

3.2 

5.6 

10 

then y = 

- .50 

- .25 

0.00 

.25 

.50 

.75 

1.00 


-Y - 




























































































4- 














































-.-h ■ 




































































































































































































r-f 































- 

TO 















1 









































































































































































— /I. - 














































T4 





































































































































































































































































































































- 










































X 













































0 - 


1 




g 






>_ 














c 





r 

7 




.C 









1 





L 






5 





± 





) 




K 

) 









c 

3 








i 






























































I 

_ 


























_ 




o 















































r 













































i 

r 













































f 














































-- 4 - 














































dl 














































- 





























































































There are no values of y for negative values of x. 
y is negative when x lies between 0 and 1. 
y increases as x increases. 












































































































































186 


ALGEBRA 


200. Most numbers are not exact powers of 10. 

Thus, from the graph on page 185, log 6 = about .78 
Correct to four decimal places log 6 = .7782. Similarly the log 

60 is 1.7782. 

The integral part of a logarithm is called the characteristic 
and the decimal part the mantissa. 

Thus, the characteristic of log 60 is 1, and the mantissa is .7782. 

201. Finding the characteristic of the logarithm of a number 
greater than 1. 

It is known that 3.53 = log 3.53 = .5478 

35.3 = 10 X 3.53 = 10 X lO-^^^^ = ... 35 3 = ^ 5473 

353 = 10 X 35.3 = 10 X IO^-^^ts ^ ... 353 ^ 2.5478 

In this last line, 353 has three figures to the left of the decimal 
point; its logarithm has characteristic 2, which is 1 less than 3. 

Rule. The characteristic of the common logarithm of a number 
greater than 1 is one less than the number of significant figures 
to the left of the decimal 'point. 

Thus, the characteristic of log 357.83 is 2; of log 70390 is 4. 

202. Finding the characteristic of the logarithm of a number 
less than 1. 

Q in.5478 

.353 = ^ = 10-5478-1 ... log 353 = 5478 _ 1 

-in. 5478-1 

.0353 = ^ = - Jq - = 10-5478-2 ... log .0353 = .5478 - 2 

1 0-5478-2 

.00353 = ■ = 10-5478-3 log .00353 = .5478 - 3 

Observing you will find that the following rule is correct. 
Rule. The characteristic of the common logarithm of a {positive) 
number less than 1 is negative; numerically it is one more than 
the number of zeros between the decimal point and the first sig¬ 
nificant figure. 





LOGARITHMS 


187 


203. The method of writing a negative characteristic. 

In §202 log .353 = .5478 — 1 . Actually, therefore, log .353 
is — .4522, a negative number. However, the positive man¬ 
tissa and the negative characteristics are retained, as follows. 

I .5478 — 1 is written: 9.5478 — 10 . Numerically the two 
expressions have equal value. Note that 9 — 10 = — 1 . 

In general, decide upon the characteristic hy the rule in § 202; 
then, if it is — 1, write it 9 — 10; if — 2, write it 8 — 10; 
etc.; and then omit the — 10 , usually. 

Thus, log .02 is .3010 - 2, or 8.3010 - 10, or 8.3010. 

Note. The negative characteristic is often written thus: log .02 
= 2.3010; again, log .353 = 1.5478. The minus sign is written over 
.the characteristic to indicate that it alone is negative, the mantissa 
being positive. Your teacher will decide which of these two methods 
you will use. 

EXERCISE 112 

What is the characteristic of the logarithm of: 


1 . 59 

6 . 72,860 

9. 5.08 

13. 

984.2 

2 . 540 

6. 11.2 

10 . 3002 

14. 

87,600 

3. 4000 

7. 367.2 

11 . 21.67 

16. 

2.193 

4. 8 

8. 50900 

12 . 100.5 

16. 

1,000,000 


Tell the number of significant figures preceding the decimal 
point when the characteristic of the logarithm is: 

17. 5 18. 3 19. 0 20. 1 21. 4 22. 2 

Write in two ways the characteristic of the logarithm of: 

23. .5 26. .07 27. .6432 29. .1007 

24. .004 26. .01003 28. .04216 30. .00008 

Tell the number of zeros preceding the first significant figure 

when the characteristic of the logarithm is: 

31. — 2 32.- 4 33. — 1 34. — 5 36.-3 


188 


ALGEBRA 


204. The mantissa of the logarithm of a number. From the 
illustrations in § 201 to § 203, it is clear that the common log¬ 
arithms of all numbers having the same significant figures have the 
same mantissas. These mantissas are given in a table of log¬ 
arithms such as appears on pages 254 and 255. 

205. Finding the logarithm of a three digit number. 

Example 1. Find the logarithm of 16.8. 

Solution. 1. In the column headed “No.’’ (page 254) find 16. 
On the horizontal line opposite 16, pass over to the column headed 
by the figure 8. The mantissa 2253 found there is the required mantissa. 

2. The characteristic is 1, by the rule in § 201. 

3. .-. log 16.8 is 1.2253. 

Rule. To find the logarithm of a number of three figures: 

1. Look in the column headed “No.” {pages 254-255) for the 
first two figures of the given number. The mantissa will be found 
on the horizontal line opposite these two figures and in the column 
headed by the third figure of the given number. 

2. Prefix the characteristic according to § 201 and § 202. 

Example 2. Find log .304. 

Solution. 1. Opposite 30 in the column headed by 4 is the mantissa 
.4829. The characteristic is — 1 or 9 — 10. (§ 202 and § 203.) 

2. .-. log .304 = 9.4829 - 10, = 9.4829. 

Note. The logarithm of a number of one or two digits may be 
found by using the column headed 0. Thus the mantissa of log 8.3 
is the same as the mantissa of log 8.30; of log 9, the same as of log 900. 

EXERCISE 113 

Find the logarithm of: 


1. 

365 

6. 64 

11. 

.841 

16. 

.000834 

2. 

571 

7. 9 

12. 

.0628 

17. 

.07 

3. 

847 

8. 5.2 

13. 

.00175 

18. 

3.14 

4. 

902 

9. 43.6 

14. 

7680 

19. 

40.8 

6 . 

200 

10. 720 

16. 

25900 

20. 

.16 


LOGARITHMS 


189 


206. The logarithm of a number of more than three digits. 

Examine 1. Find log 327.5. 


Solution. 1. 
From the table 
on page 254. 


log 327 = 2.5145 

iog 327.5 = ? 
log 328 = 2.5159 


Difference 
= .0014. 


2 . Since 327.5 is between 327 and 328, its logarithm must be between 
their logarithms. An increase of one unit in the number (from 327 to 
328) produces an increase of .0014 in the mantissa. It is assumed 
therefore that an increase of .5 in the number (from 327 to 327.5) pro¬ 
duces an increase of .5 of .0014, or of .0007 in the mantissa. 


3. .-. log 327.5 = 2.5145 + .0007, or 2.5152. 

.0014 is called the tabular difference. The zeros are usually omitted. 


Example 2. Find log 34.67. 

Solution. 1. Mantissa of log 346 = 5391 
Mantissa of log 347 = 5403 

2. Tabular difference = 12. .7 X 12 = 8.4 or 8 

3. .'. mantissa for log 3467 = 5391 + 8, or 5399 

4. .-. log 34.67 = 1.5399. 

Rule. 1. Find the mantissa for the first three figures, and the 
tabular difference for that mantissa. 

2. Multiply the tabular difference by the remaining figures of 
the given number, preceded by a decimal point. 

3. Add the result of Step 2 to the mantissa obtained in Step 1, 
writing the sum correct to four places. 

4. Prefix the proper characteristics. (See Rules, page 186.) 


EXERCISE 114 


Find the logarithm of: 

1. 

342.5 

6. 

501.6 

2. 

252.1 

7. 

28.25 

3. 

865.2 

8. 

1.158 

4. 

764.4 

9. 

7.631 

6 . 

438.3 

10. 

.5842 


11. 

.04873 

16. 

1.067 

12. 

328.2 

17. 

5.243 

13. 

453.3 

18. 

3.142 

14. 

86.43 

19. 

632.4 

16. 

3.728 

20. 

82.56 


190 


ALGEBRA 


207. Finding the number when its logarithm is given. 

Note. Some teachers call this finding the antilogarithm. 

Example 1. Find the number whose logarithm is 1.6571. 
Solution. 1. Find the mantissa 6571 in the table on pages 254-255. 

2. In the column headed “ No.’’ on the line with 6571 is 45. At the 
head of the column containing 6571 is 4. Hence the number sought has 
the figures 454. 

3. The characteristic being 1, the number must have two figures to 
the left of the decimal point. .'. the number is 45.4. 

Example 2. Find the number whose logarithm is 1.3934. 
Solution. 1. The mantissa 3934 does not appear in the table. 

The next less mantissa is 3927, and the next greater is 3945. 

That is: mantissa of log 247 = 39271 Difference 1 Tabular 
mantissa of log x = 3934 / = 7. > difference 

mantissa of log 248 = 3945 J =18. 

2. The increase of 18 in the mantissa produces an increase of 1 in 
the number. We assume that the increase of 7 produces an increase of 
3 ^ or about .4 in the number. 

3. Hence the number has the figures 247.4. 

4. Since the characteristic is 1, the number is 24.74. 

Example 3. Find the antilogarithm of 9.3940. 

Solution. 1. Again x is between 247 and 248. 

2. The difference = 13. Tabular difference = 18. = .7 

3. .'. the number has the figures 247.7. 

4. Since the characteristic is 9, the number is .2477. 

EXERCISE 115 
Find the number whose logarithm is 


1. 

2.7408 

6. 

9.5969(- 10) 

11. 2.2930 

16. 

8.9194 

2. 

1.6678 

7. 

8.3429(- 10) 

12. 1.9665 

17. 

7.7978 

3. 

.4188 

8. 

3.8497 

13. 3.8598 

18. 

3.2306 

4. 

3.8983 

9. 

7.3288(- 10) 

14. 9.7606 

19. 

1.8817 

5. 

2.9417 

10. 

.1195 

15. 4.3346 

20. 

2.9986 


LOGARITHMS 


191 


The Operations with Logarithms 


208. The logarithm of a product equals the sum of the loga¬ 
rithms of the factors. 

That is; Log MN = log M + log N 

Proof 1 . Let M = 10 ^; or a: = log M 
and let N = 10^; or y = log N 

2. MN = 10^+^ 

3. log MN = X -p y; or log MN = log M + log N. 
Example. Find the value of 7.208 X .0631. 


Solution. 1. Let v = 7.208 X .0631. 

2. log y = log 7.208 + log .0631 

3. log V = 9.6578 

4. V = .6573 


log 7.208 = .8578 
log .0631 = 8.8000 - 10 
9.6578 - 10 


By the suggestion on page 68, it may be preferable to write this as 
657, since there are only three significant figures in .0631. 

At any rate, do not keep more than four significant figures. 


EXERCISE 116 

Find the following products by logarithms: 


11. 7.026 X 8059 

12. 432.4 X 1.658 

13. 93.62 X 768.7 

14. 84.75 X .036 
16. 76.85 X .0043 


1. 42.5 X 26.8 6. 239.5 X 38.4 

2. 3.89 X 72.6 7. 871.2 X 45 

3. 535 X .621 8. 1.414 X 360 

4. .342 X 2.15 9. 8.42 X .793 

5. .654 X .368 10. 6.282 X .778 

16. C = 2 irr. Find C when tt = 3.142 and r = 13. 

17. S = irrh. Find S, when tt = 3.142; r = 11; h = IQ. 

18. I = prt. Find I, when p = $2750; r = .06; t = 4.5. 

19. Log X = .4771. Find log x^. 

20. Log y = .3010. Find log 10 y. 

21. Find 84.75 X .368 X 3.14 by logarithms. 

22. Find by logarithms the value of (73.84) 2 . 




192 


ALGEBRA 


209. The logarithm of the quotient of two numbers is the 
logarithm of the dividend minus the logarithm of the divisor. 


That is: Log (M iV) = log M - log N. 

Proof 1. Let M = 10^ or x = log M 
and let N = 10^, or y = log N 

2. M N = 

3. /. log(Af-i-iV) = X — ?/; orlog(MA") = loglf — logA'. 


Example 1 . Find the value of 
Solution. 1. Let v = 215 -h 7.25 

2. .'. log V = log 215 — log 7.25 

3. .-. logy = 1.4721 

4. .-. y = 29.65 

or y = 29.6. (See p. 68.) 


215 4- 7.25. 

log 215 = 2.3324 
log 7.25 = .8603 
Subtract 1.4721 

Find y by the rule on page 190. 


Example 2. 
Solution. 1. 

2 , 

3. 


Find the value of .192 .216. 


Log y = log .192 — log .216 
log y = 9.8488 
y = .706. 


log .192 = 19.2833 - 20 
log .216 = 9.3345 - 10 
9.8488 - 10 


Note, log .192 = 9.2833 — 10. If we subtract 9.3345 from this, 
we get a negative result. Since the mantissas in the table are all pos¬ 
itive numbers, we write 9.2833 — 10 as 19.2833 — 20. 


210. Optional method for division. Observe that 
ilf A = If X so log Af A = log M -f 

The log is called the cologarithm of A. 

Now log = log 1 — log A = 0 — log A = 10 — log A — 10. 


The advantage is that cologarithms are added 
Thus: log 


log 


75 

= log 75 

= 1.8751 


Colog 26 = 


26 X 1.8 


10 - 1.4150 

- 10 


-f- colog 26 

= 8.5850 - 

10 

= 8.5850 - 

10 


4- c log 1.8 

= 9.7447 - 

10 

Colog 1.8 = 


75 




10 - .2553 - 

- 10 

26 X 1.8 


= 20.2048 - 

■ 20. 

= 9.7447 - 

10. 











LOGARITHMS 


193 


211. The logarithm of a power (or a root) of a number is the 
logarithm of the number multiplied by its exponent; that is 
log MP = p log M. 

Proof 1. Let M = 10^ 

2. = (10"=)p = 10^^ 

3. log = xp 

4. log = p log M. 

Example 1. Find by logarithms 1.04^®. 

Solution. 1. log 1.0T° = 10 log 1.04 = 10 X .0170 = .1700 

2. 1.04i« = 1.479. 

Example 2. Find by logarithms 'V^.0359. 

Solution. 1. log ^.0359 = jlog .0359, or 1(8.5551 — 10) 

2. .-. log = 1(38.5551 - 40) (See note below.) 

3. .-. log >/:0^ = 9.6387 - 10 

4. .*. = .4352. 

Note. To divide a negative logarithm, write it in such form that 

the negative part of the characteristic may be divided exactly by the 
divisor, and give — 10 as quotient. 




EXERCISE 117 



Find by logarithms the 

value of: 



1. 

335 ^ 56 

3. 

230.4 125 

6. 

3305 -J- 1.414 

2. 

483 71 

4. 

739.8 1.73 

6. 

8.964 4- 45.25 

7. 

4.16 X 32 

9. 

43.57 X .069 

11. 

3.25 X .0063 

485 

3.14 

.007 

8. 

35.2 X 1.52 

10. 

14.07 X 347 

12. 

527.8 X .069 

53.87 

18 

2.449 

13. 

323* 

16. 

■'/ 418.5 

17. 

^92.04 

14. 

4.0252 

16. 


18. 

8.9752 

19. 

^ X 3.142 X 6.52 

20 . 3.142 X 

192 















194 


ALGEBRA 


EXERCISE 118 


Use logarithms to find products, quotients, powers, etc. 

1 . By the formula I = prt. 

(а) Find 7, if p = $4250; r = 4%; ^ = 4 yr. 

( б ) Find p if 7 = $375; r = 5%; t = 3.5 yr. 

(c) Find t if I = $840; r = 6 %; and p = $2200. 

(d) Find r if 7 = $750; p = $3750; and t = 4. 

2 . z = 2 7rr^. (tt = 3.142) 

(а) Find z when r = 13; and k = 7.5. 

( б ) Find r when h = 11.2; and z = 628. 

(c) Find h when z = 964 and r = 6.5. 



(а) Find t when I = 30. 

(б) Find I when t = 2. 

4. F = irr^h. 

(а) Find V when r = 12.4 and h = 30.3. 

(б) Find h when V = 20250 and r = 15. 
(c) Find r if = 575, and V = 8550. 

5. F = \Trr^h. 

{a) Find F when r = 6 and h = 13. 

(5) Find h when r = 7.5 and F 630. 
(c) Find r when F = 725 and h = 12. 

6. >8 = = 32.16) 

(a) Find S when t = 4.5. 

(6) Find t when S = 375. 



is the formula for the amount to which 


p dollars will accumulate at r%, compounded annually for n 
years, (a) Find A if p = $450; r = 4 %; n = 8. 

(b) Find p if ^ = 20000 ; r = 4%; = 10 . 


LOGARITHMS 


195 


EXERCISE 119. CUMULATIVE REVIEW 
1 . Find the prime factors of each of the following: 


(a) 10 a — 11 


6a3 


( 6 )* x^ + 2x^-5x-Q 

(c) 9 — 12 x"^^ + 4 


(d) 

(e) 225 + SO mn-9 - 25 m? 

(/) 9(a + 6)2 - 25 

VIE 


Rationalize the denominator of — 7 =- _ 

V5 - a/3 


6 . Solve the system 


3. Solve and check the equation 

sVx + 1 + X = 9 

4. Find the roots of .2 x^ — .32 x = .4 to the nearest hun¬ 
dredth. 

-h a 6 + 6 ^ = 19 
— ab = 7 
Group your results, and check. 

6 . (a) Multiply xi — 2 xi -{- 3 hy xi — 1 . 

( 6 ) Express the result without any fractional exponents. 

7. (a) Without solving the equation, determine the nature 
of the roots of the equation 4iX^ — 7x+9 = 0. 

ib) Form the equation whose roots are — ^ and f. 

(c) For what value of k will the roots of2a:^ — + ^ = 0 

be equal? 

8. By the use of logarithms, find the value of 

(2.53) X VTE72 
,8514 


9 . A boat can go 15 miles upstream in the time that it 
requires to go 25 miles downstream. How does the rate of the 
boat in still water compare with the rate of the stream? 

f ^2 — y 

10 . Solve the system < , ^ ^ graphically. 

[X 2 y = o 

11 . Determine graphically the roots of 3 — 2 a: = 8 . 

12 . Determine graphically the roots of3x^ — 2a: = 15. 





XIV. PROGRESSIONS 


Arithmetic Progression 


212. An arithmetic progression (A. P.) is a sequence of 
numbers, called terms, each of which after the first is derived 
from the preceding by adding to it a fixed number, called the 
common difference. 

Thus, the amounts $10.06; $10.10; $10.15; ••• form an A. P. These 
are the monthly payments for a washing machine, with interest at 6 %. 
Each payment, clearly, is 5^ more than the preceding. 

Again, 9, 6 , 3, ••• is an A. P. The common difference is — 3. The 
next two terms are 0 and — 3. 

Note. The common difference is found by subtracting any term 
from the one following it. 


EXERCISE 120 


Determine which of the following are arithmetic progressions; 
determine the common difference and the next two terms of 
the arithmetic progressions: 


6. If, 2i 3, 3f,- 

6 . 6 f, 8.5 r, 11 r, ••• 

7. 3 a, .5 a, — 2 a, ••• 

8 . 1.05, 1.10, 1.15 - 


1 . 5, 8 , 11, 14, - 

2 . 3, 8 , 13, 18, - 

3 . 1, 5, 8 , 13, - 

4 . 32, 26, 20, 14, • 


9. c4“d, 2c4-d, 3c+d”* 

10 . 6 m -f 8 n, 4 m + 9 n, 2 ??2 + 10 ?i, 


Write the first five terms of the A. P. 

11 

12 

13 

14 

16 

in which the first term is . . . 

13 

22 

3.5 

y 

a 

and the common difference is . 

6 

- 7 

4.5 

- 5 

d 


16. Find the 15th term of the A. P. whose first term is 5, 
and common difference 7. 


196 











PROGRESSIONS 


197 


213. The nth term of an arithmetic progression. It is 

possible to determine a particular term of an arithmetic pro¬ 
gression without finding all of the preceding terms. 

Given the first term a, the common difference d, and the 
number of the term, n, of an arithmetic progression. 

Find the nth term, 1. 


Solution. 1. The progression is 


Term 1 

Term 2 

Term 3 

Term 4 

Term 10 

Term n 

a 

a T d 

a ‘2 d 

n 3 d 

... ? 

... ? 


2 . What is the 10th term? the 18th? the 35th? 

3. Similarly, the coefficient of d in the nth term is 

4. 


1 = a + {n-l)d 


Example. Find the 10th term of 8, 5, 2, •••. 

Solution. 1. In this A. P., a = 8, c? = — 3, n = 10, and Z = ? 

2. The formula is I = a -\- {n — l)d. 

3. Z = 8 + 9(- 3) = 8 - 27, or - 19. 


214. The terms of an arithmetic progression between any 
two other terms are the arithmetic means of these two terms. 

Thus, the three arithmetic means of 2 and 14 are 5, 8 , 11, since 2 , 5, 

8 , 11, 14 form an arithmetic progression. 

A single arithmetic mean of two numbers is particularly 
important. It is called the arithmetic mean of the numbers. 

When two numbers are given, any special number of arith¬ 
metic means can he inserted between them. . z 

Example. Insert five arithmetic means between 13 and — 11. 

Solution. 1. There will be an arithmetic progression of 7 terms, • 
in which a = 13, Z = — 11, and n = 7. Find d. 

2. Z = a + (n - l)cZ. .-. - 11 = 13 + 6 d, or d = - 4. 

3. The progression is: 13, 9, 5, 1, — 3, — 7, — 11. 

Check. This is an A. P. which has five terms between 13 and — 11. 











198 


ALGEBRA 


EXERCISE 121 

Find: 

1. The 10th term of 4, 10, 16, •••; also the 18th. 

2. The 12th term of 17, 14, 11, •••; also the 26th. 

3. The 16th term of — 3, — 8, — 13, •••; also the 36th. 

4. The 21st term of 2, 2.05, 2.10, •••; also the 46th. 

6. The 11th term of 1, Ij, 2, •••; also the 52d. 

6. What term of the progression 1.05, 1.10, 1.15, ••• is 2.25? 
Solution. 1. This is an A. P. in which a = 1.05, d = .05, n = ? 

and I = 2.25. The formula is Z = a + (^ — l)d. 

2. .'. 2.25 = 1.05 + (n - 1) • .05. 

(Complete this solution.) 

7. What term of the progression 7, 4, 1, ••• is — 80? 

8. What term of the progression — 51, — 43, — 35, ••• is 
205? 

9. What term of the progression f, ••• is 23|? 

10. What term of the progression 1.07, 1.14, 1.21, ••• is 3.17? 
Find the common difference: 

11. If the first term of an A. P. is 6 and the 27th term is 188. 

12. If the first term is 8 and the 21st term is 108. 

13. A man is paying for a house on the installment plan. His 
payments during the first three months are $20.00, $20.10, and 
$20.20. What will his 20th and 30th payments be? 

14. The first three payments on a washing machine are $8.00, 
$8.04, and $8.08. What will the 12th payment be? 

16. In a Christmas savings fund, the payments to be made 
for 50 weeks are 5^, lOj^, 15^, etc. What will the 40th payment 
be? the 50th? 

16. Another savings fund plan calls for payments for 50 
weeks as follows: $5.00, $4.90, $4.80, etc. What will the 25th 
payment be? the 50th? 


PROGRESSIONS 


199 


EXERCISE 122 

1 . Insert three arithmetic means between 1 and 17. 

2. Insert four arithmetic means between — 14 and 16. 

3. Insert seven arithmetic means between 8 and 20. 

4. Insert five arithmetic means between 1| and 6. 

5. Insert four arithmetic means between — f and — 5. 

6. Find the arithmetic mean of 5 and 13. 

7. Find the arithmetic mean of {y — 4) and (?/ + 4). 

8. Find the arithmetic mean of Vs and V27. 

9. Find the arithmetic mean of a and h. From the result, 
make a rule for finding the arithmetic mean of any two numbers. 

Note. The arithmetic mean of two numbers is also called their 
average. 

10. (a) Find the arithmetic mean of 11 and 12. 

(6) Check your results by finding this arithmetic mean by 
use of the formula derived in Example 9. 

11. Find the common difference if three arithmetic means 
are inserted between g and h. 

12. Find the common difference if r arithmetic means are 
inserted between 5 and 45. 

13. A lot line is 165 f^et long. Fencing it, a man wants posts 
about 12 feet apart. 

(а) Not counting the end posts, how many posts would he 
place? 

(б) Exactly how far apart should he place them, to have them 
equally spaced? 

14. An orchardist is placing trees in a line which is 485 feet 
long. He wants the trees about 25 feet apart. 

(a) How many trees will he place between the two end trees. 

(b) How far apart should the trees be so that they will be 
equally spaced in the row? 


200 


ALGEBRA 


215. The sum of the first n terms of an arithmetic progression 

can be found without writing down all the terms. 

Given the first term, a, the number of terms, 7i, and the nth. 
term, 1. 


Fmd the sum of these n terms, S. 


Solution. 1. 15 = 0-1- (o -f- d) -1- (o -j- 2 d) ••• -]- (J — 2d) {I —d) -t- Z. 

2. Writing the terms in reverse order, 

aS = Z -j- (Z — d) {I — 2 d) ••• -f- (o -j- 2 d) -j- (o -|- d) -J- o. 

3. Adding the equations in Steps 1 and 2, 

2 a 5 = (o -1- Z) “h (o -|- Z) (a -f- Z) ••• -|- (o “b Z) -h (o -b Z) -f- (o -b Z). 

4. Since there were n terms in equations 1 and 2, and since a sum 
(a -b Z) results in Step 3 from each term of equation 1, 

2 y5 = 71 (o “b Z). 


5. 

6 . 


S = 5(a + !) 

By § 213, Z = a -b {n — l)d. Substituting this value of Z in Step 5, 
S = + (a + (n - l)dn. 


7. 

Note. 


S = ^ 12 a + (n - l)d] 

The formulas in Steps 5 and 7 should be memorized. 


Example. Find the sum of the first 12 terms of the pro¬ 
gression 8, 5, 2, ••• 

Solution. 1. a = S; d = — S; n = 12; S = ? 

2. S = '^{2a+ (n- l)d}. .*. >5 = 6{16 -b 11 • (- 3)}. 

3. .*. S = 6(- 17).= - 102. 


Find the sum of: 


EXERCISE 123 


1. 14 terms of 4, 9, 14. 

2. 18 terms of 6, 1, — 4. 

3. 15 terms of — 80, — 71, — 62. 

4. 12 terms of S1.08, SI.16, $1.24. 




PROGRESSIONS 


201 


Find the sum of the terms of an arithmetic progression if: 

5. The number is 16, the first is 3, and the last is 48. 

6 . The number is 39, the first is 0 , and the last is 50. 

7. The number is 19, the first is 24, and the last is — 48. 

8 . The number is 6 , the first is — y, and the last is 

9. Find the sum of the even numbers from 2 to 50. 

10. Find the sum of the numbers 3, 6 , 9, ••• 99. 

11 . Find the sum of the odd numbers from 1 to 199. 

12 . A pile of fence posts has 40 in the first layer, 39 in the 
second, 38 in the third. There are 20 layers. How many 
posts are there in the pile? 

13. In a certain school system, a teacher is paid $950 for her 
first year’s work, and is given an increase of $50 per year each 
year thereafter. What will be the teacher’s total income during 
ten years of service? 

14. It has been learned that if a marble, placed in a groove 
on an inclined plane, passes over a distance D in one second, 
then in the second second it will pass over the distance 3 D, 
in the third, over the distance 5 H, etc. Over what distance 
will it pass in the 10 th second? in the tth second? 

15. Through what total distance does the marble in Example 
14 pass in 5 seconds? in 10 seconds? in t seconds? 

16. Experiment has shown that an object will fall during 
successive seconds the following distances: 

1st second, 16.08 ft.; 3d second, 80.40 ft.; 

2d second, 48.24 ft.; 4th second, 112.56 ft. 

Find the distance through which the object will fall during 
the 7 th second; the Rh second. 

17. Find the total distance through which the object in 
Example 16 falls in 5 seconds; in ^ seconds. 


202 


ALGEBRA 


216. In an arithmetic progression, there are five elements, 
a, d, I, n, S. Two independent formulas connect these elements, 
the formula for the sum and the formula for the term 1. Hence, 
if any three of the elements are known, the other two may be 
found. 

Example 1. Given a = — f, n = 20, S = — find d 
and 1. 

Solution. 

1. S = |(a + 1). - I = 10^ - I + zj; whence Z = 

2. Z = o + (n — l)cZ. i = — I + 19 • cZ; whence cZ = i- 

Z o O 

Example 2. Given a = 7, d = 4, *S = 403; find n and Z. 

Solution. 

1. S = '^{2a+ (n-l)d\. 403 = 2|14 4. (n- 1) . 4|. 

2. 808 = » |4 » + 10); 4 n* + 10 » - 806 = 0; 

2 + 5 n — 403 = 0. 

3. .-. n - ~ ^ >/25 + 3224 _ - 5 ± V3249 - 5 ± 57 

4 4 4 ’ 

:.n=- or + 13. 

Since n, the number of terms, must be an integer, n must be 13, and 
not - 

4. Z = 7 + 12 • 4, or 55. 

Note. A negative or a fractional value of n must be rejected, 
together with all other values depending upon it. Why? 

Example 3. The 6th term of an arithmetic progression is 
10 and the 16th term is 40. Find the 10th term. 

Solution. 1. By the formula Z = a + (n — l)cZ: 

o + 5 d = 10. 
d + 15 cZ = 40. 

2. Solving the system of equations in Step 1, d = 3 and a = - 5. 

3. The 10th term: Z = - 5 + 9 • 3 = - 5 + 27, or 22. 






PROGRESSIONS 


203 


EXERCISE 124 

1 . Given I = 57, d — 4c, n = 17; find a and S. 

2 . Given a = — 7; I = 123, n = 27; find d and S. 

3. Given a = S, I = 30, S = 165; find n and d. 

4. Given a — 3, I = — 27, d = — 2 ; find n and S. 

5. Given a = 2, n — 16, ^ = 232; find I and d. 

6 . Given n = 19, s = — d = J; find a and 1. 

7. Given d = 3, = 10 , *S = 126; find a and 1. 

8 . Given a = d = — 5 = — 2 ; find n and 1. 

9. Given / = 60, *S = 240, n = 10; find a and d. 

10 . Given I = 85, d = 5, S = 750; find a and n. 

11. Given a, n, and S; derive formulas for I and d. 

12 . Given d, n, and S; derive a formula for a. 

13. Given a, I, and n; derive a formula for d. 

14. The 6 th term of an arithmetic progression is 25 and the 
11 th term is 5. Find the 30th term. 

15. The 5th term of an arithmetic progression is ys-j f^e 21 st 
term is and the last term is Find the number of the 
terms. 

16. The sum of the 2 d and the 9th terms of an arithmetic 
progression is — 8 ; and the sum of the 5th and the 10th terms 
is — %. Find the first term. 

17. Find four numbers in arithmetic progression such that 
the sum of the first two shall be — 51, and the sum of the last 
two shall be 9. 

18. Find five numbers in arithmetic progression such that 
the sum of the second, third, and fifth shall be 7; and such 
that the product of the first and fourth shall be — 14. 

19. Find three integers in arithmetic progression such that 
their sum shall be 6 , and their product — 192. 


204 


ALGEBRA 


Geometric Progression 

217. A geometric progression (G. P.) is a sequence of numbers 
called terms, each of which, after the first, is derived by mul¬ 
tiplying the preceding term by a fixed number called the ratio. 
The ratio therefore equals any term divided by the one pre¬ 
ceding it. 

Thus 2 , 6 , 18, 54, ••• is a geometric progression. The ratio is 3. 

Again, 15, — 5, + f, — is a G. P. The ratio is — The 
next two terms are + ^ and — 

EXERCISE 125 

Are the following geometric progressions? 

1. 2, 4, 8, 16 - 3. 81, 30, 21 

2. 64, 32, 16 - 4. 3, - 6, 12, - 24. 

Write the first five terms of the G. P. in which: 



5 

6 

7 

8 

9 

The first term is . . 

- 3 

72 

1 

T 

iy 

a 

The ratio is ... . 

- 2 

1 

T 

3 

1 

2 

r 


10 . If the ratio is more than + 1 , then the terms become 
(?) 

218. The nth term of a geometric progression can be found 
without determining all the preceding terms. 

Given the first term, a; the ratio, r; and the number of terms, 
n, of a geometric progression. 

Find the nth term, L 

Solution. 1 . The progression is o, ar, ar^, ••• 

2. The exponent of r in each term is 1 less than the number of the 
term. Hence the 8 th term would be ar^ and the 20th term, 

3. Similarly, the nth term must be ar^~^ 

I = ar"-i 











PROGRESSIONS 


205 


EXERCISE 126 

1 . Find the 8th term of 1, 2, 4, ^ ' 

2 . Find the 6th term of 5, 3, f, ••• 

3. Find the 11th term of 4, — 12, 36, ••• 

4. Find the 7th term of — f, 3, — 6, ••• 

6 . Find the 9th term of 5, f, -f, ••• 

6 . Find the 10th term of 

128 64’ 32 

7. Indicate the 12th term of 1, (a + 5), (a + 6)^ ••• 

8 . Indicate the 16th term of iy •*' •.; also the tth. term. 

9. Indicate the 12th term of a:, —, —, —, also the 

4 16 64 

(n + l)th term. 

10. What term of the progression 5, 10, 20, 40, ... is 640 ? 

11 . What term of the progression 7, 14, 28, ... is 448? 

12. What term of the progression 12, 6, 3, ... is -^? 

13. If the first term of a geometric progression is 4 and the 
5th term is ■^, what is the ratio ? 

Find the ratio of the geometric progression if: 

14. The first term is 4 and the 5th term is 

16. The first term is f and the 6th term is 54. 

16. The first term is 64 and the 10th term is 

17. The first term is 2 and the 7th term is 1458. 

Find by logarithms the value of: 

18. The 15th term of the G. P. of which a = 25 and r = 4. 

19. The 20th term of the G. P. of which a = 1 and r = 1.04. 

20 . The 30th term of the G. P. of which a = 50 and r = 1.02„ 

21. Find r, to hundredths, if a = 5, n = 10, and I = 50. 

22 . Find r, to hundredths, if a = 2, = 20, and I = 100, 



206 


ALGEBRA 


219. The terms of a geometric progression between any two 
other terms are called the geometric means of those two terms. 

Thus, the three geometric means of 2 and 162 are 6, 18, and 54, 
since 2, 6, 18, 54, 162 form a geometric progression. 

A single geometric mean of two numbers is particularly im¬ 
portant. It is called the geometric mean of the numbers. 

When two numbers are given, any specified number of geo¬ 
metric means may be inserted between them. 

Example. Insert three real geometric means between 9 
and -3^. 

Solviion. 1. There results a geometric progression of 5 terms, in 
which a = 9, I = and n = 5. Find r. 

2. I = .'. -3^ =_9 • r^, or 

••• r = = =t f. 

3. The progression is: 9, 6, 4, f, or 9, — 6, 4, — f, 

-Note. The other two fourth roots of if are imaginary, and imagi¬ 
nary means result from using them. 

Check. Each G. P. has three terms between 9 and 

EXERCISE 127 

1 . Insert 4 geometric means between 1 and 243. 

2 . Insert 5 geometric means between 1 and 64. 

3. Insert 2 geometric means between and 4. 

4. Find the geometric mean of 9 and 81. 

6 . Find the geometric mean of 4 y and 16 

6 . Find the geometric mean of 4 w and n. 

7 2 

7. Find the geometric mean of % and —. 

Ir a 

8 . Find the geometric mean of x and y. 

9. Insert 3 real geometric means between 4 and 16. 

10. Insert 2 real geometric means between x and y. 

11. Insert 3 real geometric means between a and h. 

12 . Find by logarithms the ratio if 6 geometric means are 
inserted between 5 and 50. 


PROGRESSIONS 


207 


220. The sum of the first n terms of a geometric progression 

can be found without writing down the n terms. 

Given the first term, a, the ratio, r, and the number of terms, 
n, of a geometric progression. 

Find the sum of the n terms, S. 

Solution. 1. S = a ar ar ^+ ar^~^ 

2. Mr rS = ar ••• + + ar” 

3. S — rS = a — ar” 

^ (1 — r)S = a — ar^. 


5. D(i_r) 



6. Since I = ar” rl = ar^ 



Example. Find the sum of the first 6 terms of 2, 6, 18, 


Solution. 1. 

2 . S= ^~ 


a = 2, r = S, n = Q. Find S. 
1-3 - 2 


= 728. 


EXERCISE 128 

Find the sum of the first: 

1 . Seven terms of the progression 6, 12, 24, ••• 

2 . Eight terms of the progression 20, 10, 5, • • • 

3. Six terms of the progression 4, — 12, +36, ••• 

4 . Five terms of the progression -^q, — h 

6. Five terms of the progression — 4, 20, — 100, ••• 

6. Twelve terms of the progression 1, x^, a:^, x^, ••• 

7. Fourteen terms of the progression 5, 5 b^, 5 b\ ••• 

8. Find the sum of 20 terms of 1, (1 + r), (1 + r)% 
*9. Find the sum of the first 10 powers of 2. 










208 


ALGEBRA 


Infinite Geometrical Progressions {Optional Topic) 

Example 1. Consider the progression 5, f, f, •••. What 
happens as the number of terms increases infinitely? 

Solution. 1. The ratio is 

2. When n = 4, I = 

c _ ^ ~ ^ ~ T ' ^ 5 ~ ^ 

" I -r 1 - i 1 - i ■ 

3. When« = 10,i = 5(J). = jg|33. 

p; _ 1 . ^ ^ 

^ a - rZ " 19,683 _ 59,049 

” 1 - r 1 - I 1 - i ' 

4. As n increases, Z decreases; also the term rl of S decreases. 

5. If n becomes infinitely large, I becomes approximately zero; there¬ 
fore rl also becomes approximately zero. 

g _ Q 

6. .'. Sn is almost -r or when n is infinitely large. 

1 — 3 ^ 


Consider now the geometric progression in which the first 
term is a and the ratio, r, is less than 1. 

The sum of n terms is 

1 — r 

When r is less that 1, r” decreases as n increases and becomes 
approximately zero as n becomes infinitely large. 

For example, the 30th power of | is Q73 74]^ g24 certainly 

almost zero. Moreover, remember this is only the 30th power of L 
And an infinitely large power, — how negligibly small it would be! 


No matter what a may be, if it is multiplied by a number 
which is approximately zero, the product also is approximately 
zero. Therefore ar” is approximately zero. 

Hence the sum of an infinite number of terms of a geometric 
progression in which r is numerically less than 1 is given by the 
formula _ a 


S = 


1 - 













PROGRESSIONS 


209 


Example 2. What is the sum of an infinite number of terms 
of 4, - f, ...? 

Solution. 1. - I ^ 4 = - I; also f) = - 

/. r = - f. 

2. the absolute value of r is which is less than 1, a = 4. 
a 4 4 12 


3. 


S = 


= y,or2.4. 


1 - r 1 + 

Example 3. What is the sum of an infinite number of terms 
of the progression 1, 




Solution. 1. 


2 . 


a = 1; r 
/. S = 


h s 


1 — r 


1 = 2. 


1 - i . 

Check. Does 1 + i + i + i + iV + ^+ -” approximately equal 2? 
{Hint. Think of these numbers as V, P', etc.) 

EXERCISE 129 

Find the sum to infinity of: 


O, 2j ■** 

2. 3, 1, i - 

3. 12, 3, f,- 

4. 5, .5, .05, 
5- 1, ric 


3 * y)\ Vi iy ■“ 

7. m, .1 m, .01 m, ••• 

8. 1, - i + h - 
9- ~ 4, — J, — 

10. 4. — 4, + -A-, ••• 


11. Find the sum of the repeating decimal .8181- 

Ol Q1 

Solution. 1. .8181 + etc. 

2 . 


81 J 1 Q « 

a = ^,andr = ^,8 = r37 


3. .*. The value is 


81 

100 


^ f 1 _ M ^ 
• r 100/ 100 


99 

100 


81 

99 


Find the value of the following repeating decimals: 


12. 

.6666 - 

16. .8333 • • 

18. 

.1111 • 

13. 

.4444 ••• 

16. .409090 ••• 

19. 

.24333 

14. 

.636363 - 

17. .3555 ••• 

20. 

.52222 







210 


ALGEBRA 


EXERCISE 130 {Optional) 

Miscellaneous Examples 

1 . Find the sum of the first 10 powers of 3. 

2 . (a) Find the sum of 10 terms of 1, 1.04, (1.04)^, (1.04)^,-*- 

( 6 ) If you have had the chapter on logarithms, find the value 

of this sum to three decimal places. 

3. If a man deposits in a savings bank $5 on the 2 nd of 
January, $10 on the 2nd of February, $20 on the 2 nd of March, 
etc.; how much will he have deposited by the end of the year. 

4. A clock strikes at each of the hours, from 1 to 12 inclusive, 
a number of times corresponding to the hour; and besides, in 
each hour, it strikes once at the end of the first quarter, twice 
at the end of the 2nd quarter, and 3 times at the end of the third 
quarter. How many times does it strike during the twelve 
hours? 

6 . (a) Findthesumof25termsofl, 1.025, (1.025)^ (1.025)®,-” 

( 6 ) By logarithms find the value of this sum. 

6 . What is the sum of the 7th to the 12 th terms, inclusive, 
of the geometric progression whose first term is 15 and whose 
ratio is ^ 2 ? 

7. On a debt of $4,000, a man was paying 7% interest. At 
the end of each year, he plans to pay $200 on the principal, and 
also the interest on the balance of the debt during that year. 

(а) Write down his payments to be made at the end of years 
one, two and three. 

( б ) How many years will it take him to pay off the debt? 

(c) How much will he have paid by the time he has freed 

himself of the debt? 

8 . How many poles will there be in a pile of telegraph poles 
if there are 25 in the first layer, 24 in the second, etc. and 1 in 
the last? 


PROGRESSIONS 


211 


9. What term of the progression 5, 10, 20, ••• is 160. 

10. Given a = i, / = , and 5 ^ in an A. P.; find d 

and n. 

11. If a man earns $360 during his first year of work, and is 
given an increase of $50 per year for each succeeding year, what 
is his salary during his 10th year, and how much has he earned 
altogether during the 10 years? 

12. A father agrees to give his son 5^ on his fifth birthday, 
lOjZ^ on his sixth, and each year up to the 21st inclusive to double 
the gift of the preceding year. How much will he have given 
him altogether after his 21st birthday? 

13. Each year a man saves half as much again as he saved 
the preceding year. If he saved $128 the first year, to what 
sum will his savings amount at the end of seven years? 

14. What is the sum of the following fifty payments: 5^, 
10|2^, 15|^, ••• etc. 

15. If at the beginning of each of 10 years a man invests $100 
at 6% simple interest, to what does the principal and interest 
amount at the end of the 10th year? 

16. Find the 10th term of the arithmetic progression whose 
first term is 7 and whose 16th term is 97. 

17. Find the arithmetic mean of a/2 and VlS. 

18. What term of the progression 3, 6, 12, 24, ••• is 384? 

19. Insert 3 geometric means between 3 and 12. 

20. A man has a debt of $3000, upon which he is paying 6% 
interest. At the end of each year he plans to pay $300 and the 
interest on the debt, which has accrued during the year. How 
much interest will he have paid when he has freed himself of 
the debt? 

21. (a) What is the arithmetic mean of 3 and 12? 

(6) What is the geometric mean of 3 and 12? 


212 


ALGEBRA 


EXERCISE 131. CHAPTER MASTERY-TEST 

1. What sort of progression is: 

(a) 75, 71, 67, 63, •••? 

(b) 100, 80, 60, 40, •••? 

2. Derive the formula for the nth term of an arithmetical 
progression. 

3. Derive the formula for the sum of the first n terms of a 
geometric progression. 

4. What is the 10th term of the progression 5, — 10, 20, •••? 

6. Find the sum of 29 terms of the progression 3, 8, 13, ••• 

6. Insert 11 arithmetic means between — 20 and + 20. 

7. If the third term of an arithmetic progression is 15 and 
the seventh term is 39, what is the first term and the common 
difference? 

8. In a savings club, the successive payments are lOjZf, 15^, 
20^, etc. for 50 weeks. How much is the total of all the pay¬ 
ments? 

9. Suppose a boy were to save 2^ during the year when he 
becomes five years old, when he becomes six years old, 8^ 
when he becomes seven years old, etc. How much would he 
have when he becomes 22 years old? 

10. What is the sum of the even numbers from 2 to 200 in¬ 
clusive? 

11. Given / = — 21; d = — 2; = 15. Find a and S. 

12. What is the sum of 15 terms of 1, (1 + r), (1 r)^, ••• 

13. What is the sum of the ten numbers beginning with 4, 
and continuing thus: 4, 8, 12, 16, •••? 

14. What is the sum of the ten numbers beginning with 4 
and continuing thus: 4, 16, 64, 256, •••? 

15. (of) What is the sum of ten terms of 1, :^, •••? 

(6) What is the sum of an infinite number of the terms? 


PROGRESSIONS 


213 


EXERCISE 132. CUMULATIVE REVIEW 

1 . Find the prime factors of each of the following: 

(а) - 2x - \ (/) a® - + a? - 1 

(б) 5 + 9 - 18 {g) x^^ + 2 x^ ^ I 

(c) 64 x^^ — {h) x^^ — 6 x^'^ + 5 x^- 

{d) 6(m — nY — {m — n) —2 (i) — 625 

(e) 2 x^ 7 X — 15 (j) (x^ — y^ — — 4 yh^ 

2 — 3v^5 

2 . Rationalize the denominator of- 7 =- 

8 + V5 


3. Find the value of each of the following expressions: 

,,, 5-2 + 3 -S" 

(o) (- 125) X ( 6 ) 


4. Determine, without solving the equation, the nature of 
the roots of each of the following equations, — giving your 
reason for your decision. 

(а) 9 — 3 a: + .25 = 0. ( 6 ) 6 a:^ — 4 x + .5 = 0. 

6 . (a) Draw the graph of the equation ?/ = 3 — 4 x. 

( б ) From the graph, determine the roots of the following 
equations, correct to tenths: 

(1) 3 x2 - 4 X = 0. (2) 3 x2 - 4 X = 2. 

6 . Determine graphically the common solutions of the 

system r ^2 _ ^2 = _ 9 

I 2 X = y + 1. 

7. If the log X = 2.7153, what is: 

(a) log 10 X ? {h) log x^ ? (c) log ? 

/- /- 3 

8. Solve and check :Vz — 6 + V 2 = / - 

Vz — 6 


9. Solve the system 


2 x2 - 3 xy = - 4. 
4 xy — 5 = 3. 


(Group your 
answers) 


10 . Find two numbers whose difference is 4, and such that 


the sum of their reciprocals is f. 







XV. THE BINOMIAL THEOREM 


221. The binomial theorem is a formula for expanding any 
power of a binomial. 

By actual multiplication: 

(a xY = o? 2 ax + x^. ( 1 ) 

(a + + 3 a^x + 3 ax^ + x^. (2) 

(a + + 4 a^x + 6 aV + 4 + x^. (3) 

Rule. To expand any power of a binomial, like (a + x)”: 

1 . The exponent of a in the first term is n and decreases by 1 
in each succeeding term until it becomes 1 in the next to the last 
term. 

2 . The first term does not contain x. The exponent of x in 
the second term is 1 and increases by 1 in each succeeding term 
until it becomes n m the last term. 

3. The coefficient of the first term 'is 1; of the second is n. 

4. If the coefficient of any term be multiplied by the exponent 
of a in that term, and the product be divided by the number of the 
term, the quotient is the coefficient of the next term. 

Note 1. Observe that the number of the terms is n + 1; that is, 
one more than the exponent of the binomial. 

Note 2. Observe also that the sum of the exponents of a and x in 
each term is n. 

Thus in (3) the sum of the exponents in each term is 4. 

Note 3. Observe that the coefficients of terms “equidistant” from 
the ends are the same. 

Thus: in (3) the 2nd and 4th terms have coefficient 4. 

Note 4. When the second term of the binomial is negative, the 
terms of the expansion are alternately positive and negative if n is a 
positive integer. 

Note 5. When the terms of the binomial are complicated mo¬ 
nomials, place each in parentheses, and afterwards simplify as in 
Example 2, Page 215. 


214 


THE BINOMIAL THEOREM 


215 


Example 1. Expand (a + xY. 

Solution. 1. The exponents of a are 5, 4, 3, 2, 1. The exponents 
of X, starting with 1 in the second term, are 1, 2, 3, 4, and 5. Writing 
the terms without the coefficients gives: 

a5 + + (?)a3a:2 + (?)aV + ('^)ax^ + x^. 

2. The coefficient of the first term is 1, and of the second term is 5 
(Rule 3). Multiplying 5, the coefficient of the second term, by 4, the 
exponent of a in the second term, and dividing by 2, the number of the 
term, gives 10, the coefficient of the third term. 

Filling in the coefficients in this manner gives: 

(a + a:)® = a® + 5 a'^x + 10 a^x^ + 10 aV + 5 ax^ + x^. 

mY 

V ' 


Example 2. Expand (2 


Here 


= 2: X = ( - -Y 



. 2 ' + 4 ( 2 )=( ~ ^ 

+ «2,ff)' + .(2,(-f+ (-=)■ 

= 16-4.8. 2 + 6-4.-^-4.2.^ + 

ryv)^ 

= 16 — 16 m + 6 + — • 


EXERCISE 133 

Expand the following: 


1. 

(m + 

6. (a — 2 hY 

11. 

(n - hY 

2. 

(x - yY 

7. (2 & + cY 

12. 

a + 

3. 

{x + 1)^ 

8. (1 + yY 

13. 

(2 .T - 3)^ 

4. 

ih - 2Y 

9. (1 - y'^Y 

14. 

(av^ + bf 

5. 

(r2 - 5^)5 

10. (a - 4 hY 

15. 

(4 + 3?Y 

16. 

(MJ 

fa bV 

(2 - 3 ) 

18. 


Write the first three terms of: 



19. 


20. (.' 2:2 + y^Y 

21. 

(a:i - yhj 


216 


ALGEBRA 


222. Writing any selected term of (a + xY. {Optional) 
If you follow the rules stated on page 214, you get; 

(a + x)^ = «”+?! 


,«-i. ^ - 1 ) 

oi . 2 


+ 


n{n — l)(?i 


1 • 2 • 3 


— x^ +•• 


Let r represent the number of any term: (such as 2, 3, 4, •••)• 
You will find that the following statements are correct: 

Rule. In the rth term: 

1 . The exponent of x is r — 1 {one less than the number of the 
term). 

2. The exponent of a is n — {r — 1), or n — r + 1. 

3. The denominator of the coefficient is 1 • 2 • 3 . (r — 1 ), 

the last factor being the same as the exponent of x. 

4. The numerator of the coefficient is n • {n — 1) ‘ {n — 2) . 

etc., until there are just as many factors as in the denominator. 

the rth term ^ n(n - l)(n - 2).(n - r + 2) ^ +i i 

(»• not 1 ) 1 • 2 • 3 . (r - 1) * ^ 

Note. You can use this formula of course. Preferably use the Rule. 
Example. What is the 4th term of (3 ah — 6)^? 

Solution. 1. (3 — 6)* = {(3 a^) + (— b)}^. 

2. the 4th term = by ( 

L • Z • 6 [ the Kule. 

3. = 56 X (243 at)(_ 6)3^ or - 13608 a%. 


EXERCISE 134 


Find the: 

1 . 5th term of (m + n)® 

2 . 4th term of {a — by 

3. 3rd term of (m + J n)^ 

4. 5th term of (5® — a‘^y 

9. Write the middle term of 


5. 4th term of (2 x^ — y) 

6 . 6 th term of (3 x — 2)^ 

7. 5th term of (| a + 2)® 

8 . 7th term of (J a: — 3)^ 








THE BINOMIAL THEOREM 


217 


223. Fractional and negative powers of (a + x). {Optional.) 

The only limitation on the use of the rules of § 221 and § 222 
when 71 is a fraction or a negative number is that a shall be 
greater than x in numerical value. This is proved in higher 
mathematics. 

Example. Expand (a + x)! to four terms. 

Solution. 1. n = 1. Use Rule, Page 214. 

2. .*. (a + x)^ 




(!- - 1 ) 


a^-23.2 _{_ 


f(f - l)(f - 2) 

1 • 2 • 3 




3 . + j a-ix + + ii r iH -*) . + 

2 2 1 1 4 * 

4. = + 3 a-ix - ^ a-ix^ + ^ a-ix^ + •••. 

EXERCISE 135 


Write the first four terms of: 

1. (a + x)i 4. {a + 2)i 

2. (1 + x)i 6. (6 - 3)i 

3. (1 — x)i 6 . (c — 4)i 


7. (1 + x)-^ 

8 . (1 - x)-^ 

9. {a + x)~^ 


224. Extraction of roots by the binomial theorem. {Optional.) 
This is of theoretical rather than practical interest. 

Example. Find ”^25 approximately to five decimal places. 
Solution. 1. The nearest perfect cube to 25 is 27. 

2 . .-. = </27 - 2 = {( 33 ) + (- 2)}i 

3. = (33)1 + K33)-t • (- 2) - |(33)-f(- 2)2 

+ A^33)-f(- 2)3 + - 


4. 

5. 


_ o 2 4 40 

3-32 9-33 81-33 

= 3 - .07407 - .00183 - .00008, or 2.92402. 


EXERCISE 136 

Find the approximate values to four decimal places of: 

1 . Vl7 . 3. 6 . ^ 7. 

2. VM 4. ^75 6 . 8 . Vm 










218 


ALGEBRA 


225. Compound interest. When interest on a sum of money 
invested for one interest period is added to the principal, and, 
with it, draws interest during the next interest period, and so 
on, then the money is said to be invested at compound interest. 

Thus, the interest on $1 at 4% for 1 year is S.04. If this is added to 
SI, making $1.04, and if $1.04 draws interest for one year, the interest 
during the second year is $.0416, and the amount at the end of two 
years is $1.04 + .0416 or $1.0816. 

This is called the compound amoimt, and $.0816 is called the com¬ 
pound interest. The interest was compounded annually at 4%. 

Problem. Find the compound amount at the end of n years 
if one dollar is invested at rate r compounded annually. 

Note. “Rate r” means 4%, or 5%, or 3|%, etc. 

Solution. 1. The interest during the first year is r, and the amount 
at the end of year 1 is 1 + r. 

2 . The interest for the second year is r(l + r). 

.‘. the amount at the end of year two is (1 + r) + r(l + r), or 

(1 + ry. 

3. The interest for the third year is r(l + r)^. 

the amount at the end of the third year is (1 + + r(l + r)^. 

This equals (1 + r)(l + rY, or (1 + rY. , 

4. Similarly, the compound amount at the end of 4 years is (1 + r)^; 
at the end of 5 years is (1 + rY', and at the end of n years is (1 + rY. 

Rule. The compound amount of P dollars invested at rate r, 
compounded annually for n years is P (l + r)^. 

Example 1. What is the compound interest on $250 invested 
at 4% compounded annually for 10 years? 

Solution. 1. In this example, r = .04 and n = 10. 

2 . .'. the compound amount of $1 = (1 + .04)^°. 

3. Log (1.04)10 = 10 • log 1.04 = 10 X .0170 = .1700 

4. .-. 1.0410 = (about) $1.48. 

5. .'. the compound amount of $1 for 10 yr. = $1.48 

6 . .•. the compound amount of $250 = 250 X $1.48, or $370.00 

7. .’. the compound interest = $370 — $250, or $120.00 
That is, about $120.00. Actually it is $120.05. 


THE BINOMIAL THEOREM 


219 


Example 2. Find the compound amount of $350 invested 
at 4% compounded semi-annually for 10 years. 

Solution. 1. Just accept the following rule: “4% compounded 
semi-annually” is secured by finding interest at “2% compounded 
annually for double the time.” 

.’. $350 at 4% compounded semi-annually for 10 years 
= 350 at 2% compounded annually for 20 years. 


2. 

A = 350(1.02)20 



3. 

log A = log 350(1.02)20 

log 350 

= 2.5441 

4. 

= log 350 -h 20 log 1.02 

20 log 1.02 

= .1720 

5. 

.-. log A = 2.7161 

adding 

2.7161 

6. 

A = $520.00 (about). 




EXERCISE 137 

Find the compound amount of: 

1 . $500 at 4% compounded annually for 10 years. 

2 . $500 at 2% compounded annually for 20 years. 

3. $1000 at 5% compounded annually for 5 years. 

4. $1000 at 3% compounded annually for 10 years. 

5. $750 at 5% compounded annually for 5 years. 

6 . $400 at 4% compounded semi-annually for 15 years. 

7. $1500 at 3% compounded semi-annually for 12 years. 

8 . $2000 at 4J% compounded annually for 10 years. 

9. $875 at 5% compounded semi-annually for 15 years. 

10 . $3200 at 4% compounded annually for 25 years. 

Note. This section gives a mere introduction to the application 
of the binomial theorem to one of the most important business applica¬ 
tions of algebra. 

The mathematics of investment and actuarial mathematics are founded 
on the mathematics taught in our chapters on exponents, logarithms, 
progressions (especially geometric), and the binomial theorem. Any 
student contemplating a course in commerce in any university will 
find knowledge of these chapters indispensable. 




220 


ALGEBRA 


226. An exponential equation is one in which the unknown 
appears as an exponent; as 10 = 4^; or 5.6 = 1.03^+^. 

Example 1. In how many years will $250 amount to $1000, 
at 4% compounded semi-annually? 


Solution. 1. Using the formula A = P(1 + r)^, 

A = $1000; P = 250; r = .02; n = 2x. 
1000 = 250(1.02)2=«^, or 4 = 1.022=>^. 

3. Taking logarithms: log 4 = 2 a; log 1.02 
log 4 .6020 


4. 


2 log 1.02 .0172 


or 35. 


EXERCISE 138 


Solve the following equations: 

1. 50 = 3^ 4. 300 = 52 ^ 7. 650 = 

2. 125 = 2y 5. 2.75 = l.U^ 8. 87.5 = 2.5^-^ 

3. 275 = 4^= 6. 24.5 = 3.2^^ 9. 9.37 = 1.05^^ 

10. In how many years will $300 amount to $500 at 4% com¬ 
pounded annually? 

11. In how many years will $350 amount to $1000 at 6% 
compounded semi-annually? 

12. In how many years will $250 double itself at 4% com¬ 
pounded semi-annually? 

13. In how many years will $400 earn $200 in compound 
interest, if it is invested at 5% compounded annually? 

14. In how many years will $100 double itself at 4% com¬ 
pounded annually? 

16. In how many years will $100 double itself at 4% com¬ 
pounded semi-annually? 

16. In how many years will $100 double itself at 4% com¬ 
pounded quarterly? 

17. In how many years will P dollars double itself at 3% 
compounded semi-annually? 




XVI. TRIGONOMETRY 


227. In applied mathematics, distances and angles often 
cannot be measured directly. In many cases, however, the 
measures can be secured indirectly by measuring other distances 
and angles, and making certain computations taught in trigo¬ 
nometry. 

Problem. Find the distance between 
A and B on opposite sides of a stream 
without crossing the stream. 

Solution. In trigonometry we learn that the 
ratio of A5 to CB in a right triangle ABC in 
which angle C is 50° is about 1.2. 

AB 

That is = 1.2 approximately. 

AB = 1.2 X 60, or AB = about 72 ft 

Observe. AB was not measured, 
it was computed after measuring 
CB and angle C. 

This relatively simple problem 
illustrates what a powerful tool of 
computation trigonometry is. 

228. Angles are measured by 
an instrument called the transit. 

This instrument contains one protractor by which vertical 
angles are measured and another by which horizontal ones 
are measured. 

Straight lines are measured by a steel tape. 

Trigonometry furnishes the means of using the measures. 

221 
















222 


ALGEBRA 


229. The tangent of an angle. In the adjoining figure, let 
Z ABC = 50°. Let each space represent 1 . PiRi is perpen¬ 
dicular to BA. PiRi, PiB, and BRi 
form triangle BPiRi, in which PiRi 
is the side opposite Z B, and BRi is 
the side adjacent to A B. 


PiRi 6 

Observe „ = - = 1.2. 

tSixi o 

.. . P 2 R 2 12 . „ 

Similarly ^ 5 - = 77 : = 1 . 2 ; 
ijXt2 lb 


and 


R3R3 


= = 1.2 

15 


By similar triangles, you can prove 

PR 



BR 


= 1.2 wherever P is on line BC, when ZB = 50°. 


This fixed value is called the tangent of ZR (tan B). 

Tan B — ~ opposite ZB in right ABPR 

BR the side adjacent to ZB in right /\BPR 

230. When an angle changes, the tangent of the angle changes. 
(a) In the adjoining figure, let 
ZCBA = 30°. 

Then tan ZCBA = tan 30° 

_AC^ 11.5 
BC 20 

{h) Z DBE = 60= 


= .57, or .6 


tan 60° = 


DE 17 


BD 10 
(c) ZFBH = 70° 


= = 1.7 


tan 70= 


= 2.75, or 2.8. 


In the table on page 257, the 








F — 


D 




c- 


tangents of many angles are given, correct to four decimal places. 
Thus: tan 30° = .5774; tan 70° = 2.7475 

Observe that the tangent increases when the angle increases. 






































































































































TRIGONOMETRY 


223 


231. The computations in trigonometry can he done by ordi¬ 
nary arithmetic, as illustrated below. They can x . 

be greatly shortened by using logarithms, if you 
have studied that topic in Chapter XIII. The 
College Entrance Examination Board expect that 
logarithms will be used. 

232. Using the table of tangents of angles. 

(See page 257.) 

(a) A problem requiring a tangent. 

In the adjoining figure BC = 25.75 ft.; angle 
B = 63° 24'; angle C = 90°. Find AC. 


Solution. 1. 


AC ^ ^ 

^ = tan Z B. 



tan 63° 20' = 1.9912 
tan 63° 30' = 2.0057 
4' = .4 of 10' 

.4 X .0145 = .0058 
.-. tan 63° 24' = 

1.9912 + .0058, or 1.9970 


2. .'. AC = BC tan 63° 24' 

3. AC = 25.75 X 1.9970 

4. .-. AC = 51.4227 or 51.42 
Z CBA is the angle of elevation of A at B. 

ZXAB is the angle of depression of J5 at A, if XA is parallel 
to BC, and angle CBA equals angle XAB. 

(b) A problem in which an angle is to be found. 

In the adjoining figure, find angle ABC. 

Solution. 1. Obviously BA = AD; and .'. BC = CD = 10 ft. 
2. tan ZB = = -5000 

tan 26° 30' = .4986 \ Diff. = 

.0014 


tan Z B 
tan 26° 40' 
.0014 ^ 7_ 

.0036 18' 

3. .-. ZB = 26° 34 


= .5000 
= .5022 


] Tab. 

\ diff. = 
.0036 


of 10' = (almost) 4'. 
lo 



EXERCISE 139 

1 . Find: (a) tan 27° 15'; (b) tan 42° 37'; (c) tan 75° 52'. 

2 . Find angle x: (a) If tan a: = .4752; (b) If tan x = 1.6842 

(c) If tan X = .8501; (d) If tan x = 1.3408. 







224 


ALGEBRA 


233. Using the logarithms of tangents of angles. (See p. 262.) 
(a) Solution of problem (a), page 223y by logarithms. 

Given BC = 25.75; ZB = 63° 24'; ZC = 90°. 

Find AC. 


Solution. 1. AC = BC tan 63° 24'. 


2. .'. log AC = log 25.75 + log 
tan 63° 24' 

log 25.75 = 1.4107" 

log tan 63° 24' = 0.3004 
1.7111. 


3. 

log 

AC = 

1.7111 


"log 

51.4 = 

1.7110 

1 Diff. 

] Tab." 

log 

AC = 

1.7111 

/ =1 

> diff. 

log 

51.5 = 

1.7118 


J =8 


1 _ 

8 — 

.125 



4. 

AC 

= 51.4125, or 

51.41. 


(See Steps 1, 2, p. 223.) 
log tan 63° 20' = 10.2991 ] Diff. 
log tan 63° 24' = ? i = 

log tan 63° 30' = 10.3023 J .0032 
.4 of .0032 = .00128, or .0013. 
Since the tangent increases when 
the angle increases, 
log tan 63° 24' = 10.2991 + .0013 
or log tan 63° 24' 

= 10.3004(- 10) 


(6) Solution of problem (b), page 223, by logarithms. 
Given AC = 5'; BC = 10'; ZG = 90°. 


Find Z B. 

Solution. 1. Tan ZB = .5000 


2. 

.'. log tan Z B 

log tan 26° 30' = 9.6977 1 Diff. ] Tab. 


= log .5000 

log tan Z R = 9.6990 J = .0013 \ diff. 

3. 

log tan Z B 

log tan 26° 40' = 9.7009 J = .0032 


= 9.6990 

.0032 X = - 3 ^ = * (^Oout) 

4. 

.*. Z 5 = 26° 34'. 

Z R = 26° 30' + 4'. 



EXERCISE 140 

1. 

Find the logarithmic tangent of : 


(a) 27° 30' 

(c) 40° 42' (c) 70° 18' 


(b) 25° 35' 

(d) 60° 56' (f) 15° 33' 


2 . Find the angle x if: 

(a) log tan x = 9.6752 (c) log tan x = 9.7913 

(5) log tan X = 9.7044 (d) log tan x = 0.0813 









TRIGONOMETRY 


225 


EXERCISE 141 

(Use logarithms or not as your teacher directs.) 

1 . In the adjoining figure, riC is perpendic¬ 
ular to CD; CD = 157.5 ft.; angle D = 60° 30'. 

Find AC. 

2 . 145.3 ft. from the foot of a high building the angle of 
elevation of the top is 39° 10 '. How high is the building? 

3. From the top of a hill known to be 185.0 ft. above the 
level of the plain, the angle of depression (See p. 223) of a house 
is 22 ° 20 '. How far away is the house from an imaginary point 
directly below the top of the hill? 

4. At a point 175.8 ft. from the foot of a building, the angle 
of elevation of the top is 51° 30'. How high is the building? 

5. From the height of 350.5 ft., the angle of depression of an 
object on the plain below is 28° 40'. Find the distance of the 
object from a point in the plain below the point of observation. 

6 . The angle of elevation of an airplane at a certain point 
P is 38° 30'. Point D, 1475 ft. distant, is directly below the 
airplane. How high is the airplane? 

7. At a point 146.8 ft. from the foot of a high chimney, 
the angle of elevation of its top is 40° 30'. How high is the 
chimney? 

8 . From a hilltop 275 ft. above the level of a lake, the 
angle of depression of one sailboat is 33° 20'. The angle of 
depression of a second sailboat directly in line with the first 
boat is 65° 40'. What is the distance between the two boats? 

9. An observer in an airplane, which is 1278 ft. high, finds 
that the angle of depression of a station on the ground is 28° 50'. 
How far distant is he from a point directly over the station? 

10 . When the angle of elevation of the sun is known to be 
37° 10', a chimney casts a shadow 125 ft. long. How high is 
the chimney? 




226 


ALGEBRA 


234. The sine of an angle. 

If you measure BPi, PiRi, BP 2 , 
P 2 R 2 , etc., you will find that 
PiRi _ P 2 R 2 _ PzRs , 

~ BP 2 ~ BPz 

Wherever P is on BC, the ratio 
of the perpendicular PR to RP al¬ 
ways has the same value. 

This ratio is called the sine of 
angle B (sin B). 



Sin B = 


PR the side opposite Z B in rt. triangle BPR 


BP the hypotenuse of rt. triangle BPR 

A table of the values of the sine for certain angles appears on 
pages 257 to 261. The sine increases when the angle increases. 


235. The cosine of an angle. (Refer to the figure above.) 

If you measure BRi, BPi, BR 2 , BP 2 , etc., you will find that 


BRi BR2 BRs 


Wherever P is on BC, if PR _L BA, 


BPi BP 2 BPz 
the ratio oi BR to BP always has the same value. 

This ratio is called the cosine oi Z B (cos B). 

^ ^ BR the side adjacent to Z B in rt. triangle BPR 

BP the hypotenuse of rt. triangle BPR 

A. table of the values of the cosine for certain angles appears 
on pages 257 to 261. The cosine decreases when the angle increases. 


236. Using the tables of sines and cosines. 


{a) Find sin 43° 52'. 
sin 43° 50' = .6926 
sin 44° = .6947 

Tab. diff. = 21 
2' = .2 of 10'. .2 X 21 = 4 
Since sine increases, 

Sin 43° 52' = .6930 


(5) Find cos 62° 18'. 
cos 62° 10' = .4669 
cos 62° 20' = .4643 
Tab. diff. = 26 
8 ' = .8 of 10'. .8 X 26 = 21 
Since cosine decreases, 

Cos 62° 10' = .4648 








































































TRIGONOMETRY 


227 


237. The logarithms of sines and cosines appear on pages 262 
to 266. If they are not used, omit this page. 

The value of the sine is always more than 0 and less than 1; 
thus sin 70° = .9397. Therefore the characteristic of the loga¬ 
rithm of a sine must be a negative number. Most of them have 
characteristic — 1, or 9 — 10. Hence in the table on pages 
262 to 266 each logarithm is printed with a characteristic 9 
(or 8) and with — 10 understood. Similarly for the logarithms 
of the cosine. 


Example 1. Find the log sin 26° 27'. 

Solution, log sin 26° 20' = 9.6470 I rp , 

1 • o Tab. din. 

log sin 26° 27' = ? \ ^ ^5 


log sin 26° 30' = 9.6495 J 
Since the sine increases when the angle increases, the 
log sin 26° 27' = 9.6470 + 18, or 9.6488. 


7' = .7 of 10' 

.7 X 25 = 17.5 
or = 18 


Example 2. Find Za:: if log sin x = 9.2892.* 

Solution, log sin 11 ° 10' = 9.2870 \ Diff. ] Tab. 

log sin a: = 9.2892 / = 22 f diff. 

log sin 11 ° 20 ' = 9.2934 J =64 

.-. Zx = 11° 10' + 4', or 11° 14'. 


99 

6l X 10' = 3.7' 
or 4' 


.3 X 14 = 4.2 
or about 4. 


Example 3. Find log cos 47° 43'. 

Solution, log cos 47° 40' = 9.8283 "j Tab. 

log cos 47° 43' = ? ^ diff. 

log cos 47° 50' = 9.8269 j = 14 
Since the cosine decreases when the angle increases, subtract the 
correction. 

.-. log cos 47° 43' = 9.8283 - 4, or 9.8279. 

Example 4. Find Za: if log cos x = 9.5253. 

Solution, log cos 70° 20' = 9.5270 1 Diff. 1 Tab. 

log cos X — 9.5253 J = 17 diff. 

log cos 70° 30' = 9.5235 J = 35 

Since Za: is between 70° 20' and 70° 30', then Za: = 70° 25'. 

* Z is the symbol for angle. 


about 5'. 




228 


ALGEBRA 


EXERCISE 142 

(Use logarithms or not as your teacher directs.) 
Omit Examples 1 and 2 if you do not use logarithms. 

(b) log sin 56° 33' 
(d) log cos 71° 44' 
A 


1 . 


2 . 


3. 


Find: (a) log sin 16° 18' 

(c) log cos 27° 46' 

Find Zx, if: 

(а) log sin x = 9.6379 

( б ) log sin X = 9.9465 

(c) log cos X = 9.9271 

(d) log cos X = 9.8053 
In a triangle XYZ, YZ is 28.50 ft., 

angle Y is 70° 15', and XF is 12.75 ft. 
Draw a triangle to represent these condi¬ 
tions and draw the altitude XW. 

(a) Find the length of XW. 

(b) Find the area. 

4. In Figure 1 , there is a right 
triangle in which AB = 10 ft., and 
BC = 18.5'. 

(a) Find angle C. (b) Find AC. 

6 . In Figure 2, OC is perpendic¬ 
ular to AB. 

(a) Find AC. (b) Find OC. 

6 . In Figure 3, A BCD is a rect¬ 
angle. 

(a) Find Z CDB. 

(b) Using ZCDB, as found, find 
DB. 

7. In Figure 4, AD is perpen¬ 
dicular to BC. 

(а) Find AD. 

( б ) Find the area of AZRC. 



Fig. 1 













TRIGONOMETRY 


229 


EXERCISE 143 

(Use logarithms or not as your teacher directs.) 

1 . At a time when the angle of elevation of the sun is known 
to be 25° 15', a chimney casts a shadow 83.25 feet long. How 
high is the chimney? 

2 . In rt. AABC, having Z A = 90°, ZB = 28° 45', and 
AB = 27.60 in., how long are: AC and RC? 

3. At a point 145.5 ft. from the foot of a tower, on which 
stands a flagpole, the angle of elevation of the top of the tower 
is 35°, and of the top of the pole 47° 25'. How high is the tower, 
and how long is the pole? 

4. A ladder 16 ft. long leans against a building. If it makes 
an angle of 57° 33' with the ground, how high on the building 
does it reach? 

6 . How long must a guy wire be to reach from the top of a 
47.5 ft. pole to a point on the ground 20.75 ft. from the foot 
of the pole, and what angle will it make with the ground? 

6 . From a window 35.33 ft. above ground level, the angle of 
depression of an object on the ground is 41° 18'. How far is 
the object from a point on the ground below the observer? 

7. Charles, on one side of a stream, holds a 9.8 ft. rod in 
vertical position, with one end on the ground. Edward, on the 
other side, measures and finds the angle of elevation of the top 
of the rod is 16° 45'. How far apart are the two boys? 

8 . A hill has a slope of about 13° 50'. From the foot of the 
hill to its top is 350.3 ft. How high is the hill? 

9. From 500 ft. in the air the angle of depression of a point 
on the ground is 37° 23'. How far is the point from directly 
below the observer? 

10 . In triangle ABC, AB = 22.75 in., Z R is 42° 33', and 
BC is 24.4 in. Find the area of the triangle. 


230 


ALGEBRA 


11 . At 119.5 ft. from the foot of a tower, the angle of elevation 
of the top was 64^ 30'. How high was the tower if the telescope 
of the transit was 5 ft. above ground level? 

12 . If the angle of elevation of the sun is 70° 15', what is 
the height of a pole which casts a shadow 17.83 ft. long? 

13. In a figure like that for Example 5, page 228, how long is 
OA, if AC is 7.5 in. and angle AOC is 22|°? 

14. Repeat Example 13, if angle AOC is 18°. 

16. In a rectangle RSTW, RS = 23.25 in., and ST = 14.2 in. 

(а) How long is diagonal RT? 

(б) How large is angle RTS^ 

16. From the top of a hill which is 83.67 ft. above lake level, 
the angle of depression of a rowboat on the lake is 18° 30'. 
How far (from directly below the observer) is the boat? 

17. How long must a ladder be to reach from a point on the 
ground 17.83 ft. from the side of a building to a point 21.50 ft. 
up on the side of the building? 

18. The height of a tower was observed at two points which 
were on the same level with and in the same straight line with 
the foot of the tower. At the nearer point, the angle of ele¬ 
vation of the top of the tower was 45° 30' and at the other was 
32° 45'. If the points were 55 ft. 2 in. apart, how high was 
the tower? 

19. In a circle whose radius is 13.5 in., chord AB makes an 
angle of 38° 55' with the radius drawn to B. How long is the 
perpendicular from the center of the circle to the chord, and 
how long is the chord if this perpendicular is known to divide 
the chord into two equal parts? 

20 . In A ABC, ZB measures 110°; AR is 12.5 in. long and 
BC is 20.42 in. long. How long is the altitude drawn from A 
to side BC, and what is the area of A ARC? 


XVII. VARIATION 


238. This chapter contains a summary and slight extension 
of the subject of functional dependence as it has appeared in 
your courses in mathematics up to this point. 

You have learned: 

(a) That one number is a function of one or more other num¬ 
bers, if a definite value of it corresponds to every value of the 
other number or numbers. 

(b) That the number or numbers in an expression or formula 
which have different values during a discussion are called 
variables; and that those which have one fixed value are called 
constants. 

Thus in F = ^Trr‘^h, tt and 3 are constants, and V, r, or h may 
be variables. 

(c) That the dependence of one number upon one or more 
others can sometimes be expressed algebraically by a formula. 

Thus: A — ^ a{h -]r c) expresses explicitly the dependence of 
A on a, b, and c. 

This same formula expresses implicitly the dependence of a 
on A, b, and c. This dependence can be made explicit by 

2 A 

solving the formula for a. Then a = ^ ^ 

In other words, that a formula expresses a functional relation 
between the variables in it. 

(d) That every algebraic expression such as x, or f x, or 
2x^ — 1, etc. is a function of x. 

In particular, that ax or ax + 6 is a linear function of x] 
that ax^, or ax^ + 6a: + c is a quadratic function of x. 

231 



232 


ALGEBRA 


239. The graph of the formula A = d h. 

A 


When b = 

0 

2 

4 

6 

8 

10 

then A = 

0 

10 

20 

30 

40 

50 


a. It is clear that: 

A increases when h increases ; 
A decreases when b decreases. 

b. When b = 2, A = 10. 

When b*= 4, = 20. 

Observe that A is doubled if b is 
doubled. Similarly, if b is trebled, 
then also A is trebled. 

c. Observe that etc. 



R: 


H- 




.6 line 


123466789 10 


That is, the ratio of any value of A to the corresponding 
value of b is. always 5. This fact can be inferred directly from 

A 

the formula. For, since A = b b, then — = 5. 


We say: A varies directly as b. 

240. While any formula can be studied in this manner, all 
these facts are true for any numbers x and y satisfying an equa¬ 
tion of the form x = my, or - = m. 

y 

One variable varies directly as another if the quotient of any 
value of the one divided by the corresponding value of the other 
is constant, as x and y above. 

Then such facts as the following are true: 
a. X increases when y increases; 
h. X decreases when y decreases; 

c. X is doubled when y is doubled; 

d. X is halved when y is halved; etc. 

Example. The perimeter, p, of a square of side s is given by the 

formula p = 4 s, from which ? = 4. Therefore, p varies directly as s. 
s 
























































































VARIATION 


233 


241. A slightly different type of variation. 

Example, a. The formula for the volume of the rectan¬ 
gular solid whose base is a square of side s and whose altitude 
is 3 is F = 3 s^. 

h. Make a table of corresponding values of s and V and draw 
the graph representing V = 3 5^. 



When s = 

0 

2 

4 

6 

8 

10 

Then F = 

0 

12 

48 

108 

192 

300 


c. Observations. 

1. V increases when s in¬ 
creases, but more rapidly 
than s. 

2. When s = 2, F = 12. 
When s = 4, F = 48. 
When s is doubled, F is 
multiphed by 4. 

3 12 = 12 ^ 0 . 

■ 22 4 ’ 42 16 ' 

10 § = 108 = 3. etc. 

& 36 


to 


Therefore F is proportional 
This fact can be inferred directly from the formula. For, since 


F = 3 s2, then 


3. 


We say, F varies directly as the square of s. 


In general, one number varies directly as the square of another 
if the quotient of any value of the one divided by the square 
of the corresponding value of the other is constant. 

Example 1. Since the area. A, of a square of side, s, is given by the 

A 

formula A - s^, then — = 1. 

S2 

Therefore A varies directly as the square of s. 

This means that A is multiplied by 4 when s is doubled. 

that A is multiplied by 9 when s is trebled; etc. 





































































































234 


ALGEBRA 


242. Another type of variation is illustrated by the formula 

I = PRT. Let r = 2. Then I = 2 PR. 

Since 7=2 PR, then ^ = 2 . Hence, in general I is pro- 
HR 

portional to the product of P and R. 

We say, I varies jointly as P and R. 

In general, one number varies jointly as two or more others 
when any value of the one divided by the product of the cor¬ 
responding values of the others is constant. 

Example. Since V = Iwh, -^ = 1. Hence V varies jointly as 
Iwh 

I, w, and h. 


EXERCISE 144 

Complete the following sentences : 

A 

1 . Since A = ah, and therefore — = ?, then A varies 

ab 

_as a and b. Hence, if a alone is doubled, then A 

is_; if b alone is trebled, then A is_; 

if a is doubled and b is trebled, then A is_ 


2 . Since A = ab, if a = 2, then 


A 


Hence A varies 


If b is halved, then A is 


If A is doubled, then b is. 


A 

3. Since A = xH, then — = ? Therefore A varies as 

_Hence, if r is doubled, A is multiplied by 

-; if r is trebled, A is multiplied by_ 

4. Since 8 = 4: xH, then 8 varies_as_ 

Hence if r increases, 8 _, only_rapidly; 

if r is doubled, then 8 is_ 

6 . Since V = ^ xr^, then V varies_as__ 

Hence if r increases, V __, only_ r apidly ; 

if r is doubled, V is multiplied by_ 






















VARIATION 


235 


243. A quite different type of variation is illustrated by the 
following example. 

Example, a. In the formula A = hh, for the area of a 
rectangle, let A = 20. Then hh = 20 is the formula for all 
rectangles having an area 20. 

h. Make a table of corresponding values of h and b and draw 
the graph of the relation hb = 20. 


When b = 

1 

2 

4 

5 

10 

20 

Then h = 

20 

10 

5 

4 

2 

1 


c. Observations. 

1 . When b increases, h de¬ 

creases. 

2 . When h increases, b de¬ 

creases. 

3. When 6 = 2, A = 10. 

When b = 20, A = 1. 

So, if b is multiplied by 10, h 
is divided by 10. Similarly, if 
b is multiplied by 5, h is divided 
by 5. 

We say h is inversely proportional to b, and that b is inversely 
proportional to h. 

In general, one number varies inversely as another when the 
product of any value of the one and the corresponding value of 
the other is constant. 

Example, p = br is a well-known formula. Suppose we 
consider the case in which p is constant, — like 100, or 50, or 
80, etc. Then br — a constant. 

Hence b varies inversely as r ; that is, the base increases if 
the rate decreases, and vice versa; if the base is doubled, the 
rate is halved; etc. 


TT 

1 





















J L 

1 

1 




















or 
















































































































































tt 
































































































kf 
























IL 

r 










































































L 





















.p 


























V 

































































-r 

)- 






















0 






0 


3 












2 




n 


It 








1 














































































236 


ALGEBRA 


EXERCISE 145 

Miscellaneous Examples of Variation 

1. If E = Iwh and I is constant, then V varies- 

as w and h. If then w alone is doubled, V is_; if 

w is doubled and h is trebled, then V is_ 

2. If F = Iwh and I and h are constant, then_ 

3. If 7 = PRT, where 7, P, R, and T are variable, then 7 

varies_as_ 

4. If 7 = PRT, and R and T are constant, then 7 varies 

_as_ In this case if 7 is multiplied by 

5, then_is_by_ 

5. If 7 = PR T, and 7 and T are constant, then_ 

varies_as_In this case, if P is increased, 

then_is_ 


6. 5=-(a+/) expresses the dependence of >S on 
If a and I are constant, then S varies_as_ 


7. I = a {n — l)d expresses the dependence of I on 

_If n and d are constant, / is a function of_; 

then if_increases, /_ 

2 S 

8, From the formula in Example 6, n = 

function of_ 

creases, n _ 

creases, u _ 


so 71 IS a 


a I 

If S and / are constant, when a in- 
If a and I are constant, when S de- 


9. P = 32 + f (7 expresses the functional relation between 
corresponding Fahrenheit and Centigrade temperature readings. 
If C increases, then F _ 


10. Z = 2 TTrh is a formula in solid geometry. It indicates 

that Z varies_as_and_When r 

and h are both increased, then Z _If r is constant, 

when h is doubled, Z is_ 

































ADDITIONAL EXAMPLES FOR 
REMEDIAL PRACTICE 


EXERCISE 146 

Removing and inserting symbols of grouping 
Remove the parentheses or other symbols of grouping: 


1. 3 c + (2 a: — ?/) 

2. 2x — {y z) 

3. 4iX — {y — z) 

4. 5 a — (2 6 + c) 

5. 7 c + (- 3 a: + 2 /) 


6. 5 s — S t — w) 

7. 6 r — (4 a — 6) 

8. 2 a + (3 6 — c) 

9. z - 2 X -\- y) 

10. 6 - (2 a + 3 c) 


Remove symbols of grouping and combine terms: 


11. 3 X — (5 X — 2) 

12 . (2 a — 6) — (4 a + 6) 

13. (4 X - 5) - (3 a: - 2), 

14. (2 6 - 7) + (- 3 6 + 9) 


15. {-2x-i-y) - 2x - y) 

16. 5 x^ — (3 + x) — 4 

17. (3 r — 2 5 ) — (r + 3 s) 

18. (5 p 3 q) - 3 p + q) 


Remove all the symbols of grouping, starting with the inner 
ones: 


19. r (5 r — {2 s — t}) 

20. X - [ 2 / - (2 X + 2 /)] 

21. a — {3a+(2a — 5)} 

22. 8 — []— 2+(x + 5)]] 

23. 2/+ { - 3 2/+ (2 2/ - 1)1 


24. / + [3 - (2 / + 4)] 

26. w — (5 + [3 w — 6]) 

26. s - 3s - (2]- 5)1 

27. 9 - [4 X - (2 X - 6)] 

28. z + { - 7 - (2s - 3)1 


Inclose the last three terms of each expression in parentheses, 
preceded by the sign of the first of the terms which are inclosed: 


29. 2 /^ + x^ — 2 X + 1 
39. 2 X — 1 

31. — x^ — 2 X — 1 

32. a^ + 4: — 4x-\-4:X^ 

33. 6^ — 9 — 6 c — c* 


34. a 4- 6x — cx + dx 

35. r — sx 4- tx — wx 

36. 6 — ox — 6x + cx 

37. w — mx px qx 

38. 16 — x^ + 10 X — 25 
237 


238 


ALGEBRA 


EXERCISE 147 

Factor as taught on page 19. 


1. 

x2.+ 8 X + 15 

2. 

m2 — 11 m + 28 

3. 

18 - 9 i + ^2 

4. 

3 x2 + 7 X + 2 

5. 

6x2— 

6. 

3 x2 — 19 X + 6 

7. 

x2 + 7 X + 12 

8. 

7 2/2 + 92/4-2 

9. 

2'w^ — 7 w ^ 

10. 

24 - 11 2 + z2 

11. 

6 x2 + 7 X + 2 

12. 

7r2 + 4r — 11 

13. 

— 10 wr + 24 7*2 

14. 

5 52 + 7 5 - 6 

16. 

12 x2 - 5 X - 3 

16. 

12 ^2 - 13 ^ + 3 

17. 

C2 - 11 + 30 c^2 

18. 

2 x2 — X — 15 

19. 

9 2 /^ - 6 2 / - 8 

20. 

m2 + 12 mn + 35 'n/ 

21. 

7 a2 - 10 a + 3 

22. 

21 A^ + 2 A - 8 

23. 

0 

1 

CO 

1 

24. 

0 

rH 

1 

CO 

1 

00 

t-H 

26. 

6 a:* + 13 X + 6 

26. 

1 

(M 

1 


27. 9 - m2 

28. — 10 a^a + 25 

29. z 2 _ 4 2 _ 21 

30. — A:'if- 

31. 32 - 4 s - s 2 

32. 1 - 36 aW 

33. 2 w2 - 3 w - 20 

34. X? + 12 xy -\- 36 

36. 60 — 4 w — ^2 

36. 1 + 14 X + 49 x2 

37. 10 c2 + 9 c - 9 

38. 4 a2 - 20 a + 25 

39. 25 u2 - 1 

40. 54 + 3 a — a? 

41. 9 a2 — 6 + 52 

42. 44 + 7 ^ - ^2 

43. x2 + 2 X — 35 

44. + r2 — 2 

45. 4 — 4 a26 + 

46. 7 a2 - 26 a 6 - 8 62 

47. 9 x2 - 4 1/2 

48. 18 — 7 a — a2 

49. — 18 w + 72 

60. 4 a2 — 8 a — 21 

61. (? - cd- 20 

62. x2 — 8 XI/ — 33 2/2 


ADDITIONAL PRACTICE EXERCISES 


239 


EXERCISE 148 

Perform the indicated operations: 
4a2-4a+l 2 + a ^a^- 


2 a 


6 . 

7. 

8 . 

10 . 

12 . 

14. 

16. 

16. 

17. 

18. 

19. 


4a2 
3 + 5 
X _ y 

3 _ 4 

^ y 

a — 1 
a + 1 a — 1 


8a^ 

1 


3. 


■ 1 

3 a 


a + 1 


+ 


'a: 4- 2 


+ 


9a 
4a2 + 1 

a? - 1 

X 


8 

^^ 

1-? 

X 


5 X 

X — 3 


4 x^ -h 3 X — 1 
-f — 12 


X X — 3/ \x — 2 
ax — hx — ay -\- by 

a2 - ¥ 

a? - 2a- 3b 4 a^ - 9 a 


X -f- 3, 


2 a^ - 3 a2 
1 


a7 - 49 
1 


9. 

11 . 


2 ac — 2 he — ad + hd 
d? -4 

5x + 2 . 1 


a2 - 4 a6 + 4 62 4 62 


13. X 


2x2+x-10 x-2 

x3 + 27 


x2 + 3 X + 9 


a2 + 4 a - 60 a2 - 4 a - 12 
m — 1 m+1 m— 6 

m — 2 w+2 4 — m2 

6x - 11\ 


2X-1+ , . 

V X + 4 

Find the value of 

A x2 + 4 X — 21 

V ~ x2 + 2 X - 8 

x2 + 2 X2/ 4- ^2 _ 2:2 

x2 — 2 XZ + 22 — ^2 


X + 3 


3 X 


+ 17 \ 


X + 4 / 
1 


-4^^lWhenx = -. 

+ 1 , ±:il\ 
X — 2 X 4- 4/ 


20 . 


6a2 


8a2 


ax 4- ay — bx — by 
a2 — 2 a6 4- 62 
18 a — 5 4a2 4"6a4“2 


4 a2 - 16 a + 15 12 a2 - 5 a 


21. 


4a2 - 9 









































240 


ALGEBRA 


EXERCISE 149 

Uniform Motion Problems 

1 . A freight train runs 6 miles an hour less than a passenger 
train. It runs 80 miles in the same time that the passenger 
train runs 112 miles. Find the rate of each. 

2. A man walked 10 miles. He returned in a car at the rate of 
30 miles an hour. If he was gone 4 hours, what was his rate 
while walking? 

3. One auto, traveling at the average rate of 42 miles an hour, 
and a second, traveling at the average rate of 30 miles an hour, 
left points which are 252 miles apart and traveled toward each 
other. In how many hours will they meet? 

4. A railroad and an auto highway run beside each other for 
a considerable distance. If a train left a starting point at 
8:00, running 30 miles an hour, and an automobile left the same 
point at 8:15, running 40 miles an hour, how far from the 
starting point will the automobile overtake the train? 

6. A man traveled 100 miles at the rate of 30 miles an hour. 
At what rate must he return if he wishes to cut the time for the 
return trip to three fourths of that for the outward trip? 

6. (a) One train travels / miles an hour and a second travels 
s miles an hour. If they start toward each other from points 
which are d miles apart, leaving at the same time, in how many 
hours will they meet? 

(b) After you find the result for part (a), answer the following 
questions: 

1. The time is a function of_? 

2. If / and s are constant, then the time increases when 

the distance d _ 

3. If d and / are constant, and s increases, then the time 

4. If d is constant and / and s both increase, then_ 






ADDITIONAL PRACTICE EXERCISES 


241 


EXERCISE 150 
Mixture Problems 

1 . One bar of silver alloy is 40% silver; another bar is 25% 
silver. How many ounces of each must be taken to make 80 oz. 
of a new alloy (when melted together), which will contain 35% 
silver? 

2. How many pounds of OOj^ candy and of SOjz^ candy must 
be taken to make 50 lb. of mixed candy to sell at 40{zf per pound? 

3. Chemically pure sulphuric acid contains 95% of sulphuric 
acid. How many pints of it and of distilled water must be 
taken to make a one gallon mixture which shall be 60% pure? 

4. ‘‘Glycerine and rose water” is a mixture of pure glycerine 
and perfumed distilled water. How much water must be added 
to a pint of such a mixture which contains 50% of glycerine 
to make a new mixture which will contain 40% of glycerine? 

6. In a mixture of sand and cement containing one cubic 
yard, 40% is cement. How much cement must be added so 
that the resulting mixture will be 50% cement? 

6. Stronger ammonia water contains 25% of ammonia. How 
much water must be added to one gallon of the stronger ammonia 
water to make ammonia water which is 10% pure? 

7. Chemically pure hydrochloric acid contains 33% of hydro¬ 
chloric acid. How many pints of it and of distilled water are 
needed to make one quart of dilute hydrochloric acid which is 
10% pure? 

8. Tincture of iodine contains 6.5% of iodine. How much 
alcohol must be added to one pint of the tincture of iodine to 
make a solution which contains 3% of iodine? 

9. A normal salt solution contains 85% of salt. How much 
distilled water must be added to one quart of a normal salt 
solution to make a solution which contains 5% of salt? 


242 


ALGEBRA 


EXERCISE 151 

1. A gallon occupies 231 cubic inches of space. Express as 
a formula the number of gallons, G, which can be held by a 
tank I feet long, w feet wide, and h feet high. 

2. A cord of wood contains 128 cubic feet. Express as a 
formula the number of cords, C, in a pile of wood m feet long, 
n feet wide, and r feet high. 

3. A bushel contains 2150.4 cubic inches. Express as a 
formula the number of bushels, B, in a bin a feet long, h feet 
wide, and c feet high. 

4. a. One cubic foot contains about gallons. Express as 
a formula the number of gallons, G, in a tank r feet long, s feet 
wide, and t feet high. 

b. By the formula of part a, find the number of gallons in a 
tank 7 feet long, 3 feet wide, and 2 feet high. 

c. By the same formula, find the number of gallons in a 
tank 3 yards long, 1.5 yards wide, and 30 inches high. 

6. A bushel of potatoes occupies 1.5 cubic feet. Express by 
a formula the number of bushels, B, in a bin x feet long, y feet 
wide, and 3 feet high. 

6. a. A cubic foot of water weighs 62.5 pounds. Express by 
a formula the weight in pounds, P, of the water held by a tank 
L feet long, W feet wide, and H feet high. 

b. Express the weight in tons, T, of the water in the same 
tank. 

c. Find the number of tons of water in a tank 8 feet long, 
4 feet wide, and 3 feet high. 

7. a. Coal is 1.3 times as heavy as water. Express by a 
formula the weight in tons, T, of the coal that fills a box I feet 
long, w feet wide, and h feet high. 

b. Find the weight in tons of the coal in a box 9 feet long, 
3 feet wide, and 2 feet high. 


ADDITIONAL PRACTICE EXERCISES 


243 


EXERCISE 162 


1 . (a) Solve r = - + ^ for t. 

(6) Find t when T = 413.2 and a - 273. 

2. (a) Solve u = ) v for M. 

\M + mj 

(6) Find M when w = 40; m = 5.5; and v = 1000. 


3. (a) C = 


Kah 


Solve it for h. 


(b) Find b when C = 12,750; K = 315, and a = 8. 
c — b 

4. 5 = - r- (a) Solve it for c. 

a — b 

(b) Find c when s = 3.25, a = 25, and b = 22. 


6. I = —— (a) Solve it for t. 

(b) Find t when I = .00002, M = 90, and L = 90.2. 
6. p = ——h d. (a) Solve it for a. 


(b) Find a when p = 135, d = \, and t = \. 

7. A = P -\- PR T. (a) Solve it for T. 

(6) Find T when A = $3750, P = $3000, and R = 5%. 

8. S = ^ (a 1). (a) Solve it for n. 

(b) Find n when S = 473, a = 17, and I = 69. 

(c) Find I when S = 3275, a = 30, and n = 10. 

id) Solve for a; find a when S = 280, I = — 50, and n = 14. 

9.8 = ^ • (a) Solve it for r. 

r — 1 

(b) Find r when S = 2042, I = 1024, and a = 6. 

(c) Solve the same formula for 1. 

(d) Find I when S = 2400, r = 3, and a = 6. 

(e) Solve the same formula for a. 

(/) Find a when S = — 200, r = J; and I = 10. 







244 


ALGEBRA 


EXERCISE 153 

Solve the following systems for x and y : 

cx — dy = a 


j Sx A- ay = 3 

6. 

\2x — by = A 

(Ax — my = n 

7. 

\ X At vy = (l 

J Sax — y = b 
\ ax At 2y = c 

8. 


17. 


18. 


mx ny = 1 
cx — 2y = 2 
ax — by = S 
bx — ay = 1 

f ?^4.y = ? 

I a 6 3 

3a; y _ 1 
[a 6 6 

{Ax _3y _S 
\ a 6 2 

I = § 

[a 6 3 

m 


{^ 2y_b 
\ c d 2 
^ _ 11 
[ c d 12 

(^-^=0 

) t r 

i ^ 4 6 


10 . 


X —by = c 
j mx -\rriy = m 
\ a; + my = n 
Ax — sy = t 
sx — ry = t 
3a; — 42/ = 5 
a; + 2?/ = 6 
f 5a; — 3?/ = 1 
\2a;+5y= 19 



11 . 


{ Ax A-Zy = l 


[ 3a; — 5?/= — .7 
- 22 , = 1.2 


13. 


27. < 


[ 3x + 41/ = .2 
f ax —by = a 
\bx-\- ay b 
J ax — by = c 
\bx — ay = c 
j mx A-ny = d 
\nx — my = d 

I 2"’" 3 6 

I = i 

[3 6 

! 7 14 ^ 


28. < 


29. 


30. 


2 + 6 

?-? = _ 

X y 

5 _ 6 _ 6 
I a; y 2 

a b 

_ 1 

. a 6 
' X — c ^ d — y 
d 2c 

^=3 


ADDITIONAL PRACTICE EXERCISES 


245 


EXERCISE 154 

1. Solve the equation Vx — 2 + V'2x-i-5 = 3. 

Solution. 1. Vx — 2 + V2x + 5 = 3. 

2. V2 a; + 5 = 3 - V a; - 2. 

3. Squaring, 2 a; + 5 = 9 — 6 V^ — 2 + a; — 2. 

(Complete as in Examples 1-6.) 

Note. You will get two apparent roots, but one will not check. 
Remember that the radical indicates the principal root. 

2. "Vy + 5 — 'Vy = 1 3. — 3— l = — 10 

4. Vs m - 11 + Vs m + 10 = 7 

5. V?/4-7—V?/ — 5 = 2 

6. Vx + 15 - Vx - 24 = 13 

7 . Vc + 20 - Vc - 1 = 3 8. V4 2 / - 40 + 14 = 10 

9. 2 V 3 X - 2 - sVx - 2 = 2 
10. Solve the equation Vt/H-S—V i/4-8 = — V'y. 

Solution. 1. Squaring both members: 

y + 3 - 2V(y + 3)(j; + 8) + 2/ + 8 = 3/. 

(Complete the solution as in Examples 1-6.) 


11. 

Vx + 2 - 

1 

to 

11 

V2x 

12. 

V2t+ 1 -- 

= 2Vt - Vt- 3 

13. 

Vx + 5 — 

II 

1 

< 

CO 

to 

14. 

Vs - 2 c 

- V7 + 6 c 

' = Vl + 2 

15. 

V m + 3 -f 

- ^3 Vm + 3 

16. 

Vx +2 

V X 

2 

Vx 

Vx +2 

sVx + 2 


9 + V^ _ n Vsx + 1 - Vsx 

9 — 2 V X ^ VSx+l + VSx 


17. 


13 


















































246 


ALGEBRA 


EXERCISE 156 

Solve by completing the square: 

Example. Solve for x the equation + 3 aa: + 4 6 = 0. 
Solution. 1. 3 ax = — 4: h. 

2 . 


iof 3o = 


m- 


9a^ 

4 


3. A/9 g2 


(^) 


9 a2 9 

x^ 3 ax -\—T— = —i-4 0 . 


4. 

5. 

6 . 


a: + 


3 a 


9 a2 _ 16 6 


,3a \/9 a2 - 16 & 

^ ^ *-2- 


3a, V9 a^ — IQb — 3 a ± V9 a^ 


16 6 


_ — 3 a + \/9 a^ — 16 6 —3a— V9 a^ — 16 b 

7. xi = -s-; X2 = - 


Check: Does 
/ — 3 a + y/9 a^ ■ 


16 6 


3a— V9 a^ ■ 


2 

■ 16 T\ 


Does I 

Does 

Does 


- 3 a + V9 a2 - 16 6^ ^ - 3 a - V9 a^ - 16 b 




2 

- 3 a)2 - 


j — — 3 a? Yes. 

)- 


+ 4 6? 


(9a2 - 16 6)\ 


9 a2 - 9 a2 + 16 6 


+ 4 6? 


1. a:^ + 2 aa: + c = 0 

2. + 4 6a: ■+ ^ = 0 

3. x^ — 6 bx — c = 0 

4. 2a:^ — 4a: + m = 0 

5. 2a:^ — 2a’ — w = 0 

6 . 3 a:^ — 6 aa: + 6 = 0 

7. a:^ + 2 aa: — 6 = 0 


= 4 6? Yes. 

8. a:^ — 2 mx = 1 + 2 w 

9. a:^ + 6 a: = — 6 t 

10. — 2 aa: = 6^ + 2 a6 

11. ax^ — 2 X + 1 = 0 

12. ax^ + X + 1 = 0 

13. ax2 — 2 c^x + m = 0 

14. +x2 + Rx + G = 0 
























ADDITIONAL PRACTICE EXERCISES 


247 


EXERCISE 156. QUADRATIC EQUATIONS 
Solve these equations by factoring when that is possible; 
otherwise complete the square or use the formula. 

_ ^ — X „ X 


A 4 . = _ 15 


_6_ . A 

7 - 2 / 2 / 3 


= 2 - 


a: — 2 2 a: — 6 

3 y 5 2 / - 4 
4: - 5y Sy 

X 


= 7. 


a — 6 
3 


X + 4 

2 _ 2a - 9 

a — 6 


a — 5 


1 6 ^ a: 


9. 


10 . 


11 . 


12 . 


_ 8 
3 

a:^ + a: 


8. i = 

2 ^ - 2 


+ 


7+ 3t 


X — 1 
X 


x^ + 2x - 2 


a:4-2 a:^H-5a:4-6 a: + 3 

2 2 / + 1 y - 9 


7-2/ 
1 


= 1 


+ 


1-32/ 

7 a: 26 


a: — 2 a: + 2 a:^ — 4 


13. 

14. 

15. 

16. 

17. 

18. 


4 a: — 3 

5 5 

5 


3 — r 
8 

a — 3 
4 


+ 4 = - 

_ y ~ ^ 

y - 4 

+ 


2 a: 


4 a: 


6 — r 
a 

a — 4 


3 — r 
a — 6 


a 


2c _ _ 9 

c - 2 12 


19. 


20 . 


21 . 


22 . 


23. 


24. 


W 


W - 1 

a: — 1 


5W 

12 


2a: 


y - 2 
X — 3 


X + 4 
2 / , 4 

2 / + 2'^3 


= — + - 
X — 2 4 X 

5 3x - 16 


5 — X 
6 m 


X - 

+ 5 = 


■ 8 

3 — 4 m 


3 















































248 


ALGEBRA 


EXERCISE 157 

Solve for x, using any convenient method: 


1 . — 4 ca: — 5 = 0 

2 . 3 a:^ — rx — 2 = 0 

3. 4- 2 xp + = 0 

4. 3 x^ + 2 mx — 71 = 0 
6. 5 = ax + i gx^ 


6. x^ — Pa: 4- (P — 1) = 0 

7. x^ 4- (a 4- l)x 4- a = 0 

8. x^ — (a — h)x — ah = 0 

9. cx^ — (c 4- d)x 4- d = 0 
10. TTx^ 4- TTxl — s = 0 


11. (a) The square of a certain number equals the sum of 
that number and n. What is the number? 

(b) Using your results as formulas, find the number, when 
71 = 12; also when n = 10. 

12. (a) If the square of a certain number be increased by n 
times the number, the result is m. What is the number? 

(h) Using your results as formulas, what is the number when 
n = S and tti = 9? also when n = 5 and m = 10? 

13. (a) There are two consecutive even integers whose 
product is p. What are they? 

(b) Using your results as formulas, find the integers when 
p is 48; also when p = 80. 

(c) Suppose p = 9. Do the results have real meaning accord¬ 
ing to the statement of the problem? 

14. (a) The length of a certain rectangle exceeds its width by 
r feet. Its area is s square feet. What are its dimensions? 

(6) Using your results as formulas, find the dimensions when 
r = 12 and s = 13. 

(c) Also find the dimensions when r = 10 and s = 100. 

15. (a) The sum of the base and altitude of a certain rec¬ 
tangle is p feet, and the area of it is A square feet. What are its 
dimensions? 

(6) What are the dimensions when p = 40 and A = 375? 


ADDITIONAL PRACTICE EXERCISES 


249 


EXERCISE 158 


Find: 

1. m so that the roots of - 14 mo: + 9 = 0 shall be equal. 

2. c so that the roots of 2/^ — 5 ?/ = 8 c shall be equal. 

3. a so that the roots of - 3 a: + 4 = 0 shall be equal. 

4. Determine n so that 2 — a/s shall be a root of the equa¬ 
tion 5 X -h n = 0. 


6. Determine p so that V2 - 1 shall be a root of the equa¬ 
tion 2 x"^ — px 4 = 0. 

6. Without solving the equation or substituting in it, de¬ 
termine whether 1 + V2 and 1 - V2 are the roots of the 
equation x^ — 2 x — 1 = 0. 

7. Without solving or substituting in the equation, determine 
whether 3 — 2 i\^2 and 3-1-2 zV2 are the roots of the equation 
2:2 - 6 a: -f 17 = 0. 

8. Find the value of m so that the sum of the roots of the 
equation 3 mx"^ -f (8 m — l)a:-|-7 = 0 shall be 3. 

9. Find the value of c in the equation 2 2:2 -f (2 c— 1 ) 2 : 
+ 10 c2 = 0 so that the product of the roots shall be 20. 

10. Find the value of k so that the roots shall be equal in 
the equation kx^ -|- (^ — 1 ) 2 : — 1 = 0. 

11. Find the value of t so that the roots shall be equal in 
the equation ^ 2:2 -f- {2t — l)x — 2 = 0. 

12. Find the value of t so that one root shall be double the 
other in the equation ^ 2:2 -f (2 f — 1 ) 2 : — 2 = 0. 

13. What relation must exist between k and c in order that 
the roots of kx^ — 3 2 : -f- c = 0 shall be equal? 

14. In the equation 2:2 -f- 2 /i: 2 : -h 3 = 0, what must the value 
of k be so that one root shall be 2 more than the other? 


16. What is the value of ^ in 2:2 -|- — 4)x — 5 = 0 if the 

roots are equal in value but opposite in sign? 


INDEX * 


Abscissa, 64. 

Absolute value, 2. 

Algebraic expression, 3; value of 
an, 3. 

Angle, of elevation, 223; of de¬ 
pression, 223. 

Arithmetic, mean, 197; progres¬ 
sion, 196. 

Ascending powers, 9. 

Axis, horizontal, 64; vertical, 64. 

Base, 4. 

Binomial, 5; square of a, 18; 
theorem, 214. 

Cancellation, in an equation, 48. 

Changing signs, in an equation, 48; 
in a fraction, 32. 

Characteristic, 186. 

Circle, equation of a, 147. 

Clearing of fractions, 49, 50. 

Coefficient, 4; numerical, 4. 

Common, difference, 196; loga¬ 
rithm, 185. 

Complex, fraction, 40; number, 
133. 

Computation, accuracy in, 68. 

Conditional equation, 46. 

Conjugate, complex numbers, 135; 
surds, 109. 

Constant, 61. 

Coordinates, 64. 

Cosine, 226. 

Degree, of an equation, 46, 116. 

Denominator, 30. 


Descending powers, 9. 

Determinant, 90. 

Difference, common, 196. 

Discriminant, 139. 

Division, synthetic, 166. 

Elimination, by addition or sub¬ 
traction, 80; by substitution, 82. 

Ellipse, 148. 

Equation, 46; cancelling terms in 
an, 48; changing signs in an, 
48; complete quadratic, 116; 
conditional, 46; identical, 46; 
incomplete quadratic, 116; in¬ 
tegral, 77; linear, 55; members 
of an, 46; of first degree, 46, 77; 
quadratic, 116; radical, 158; 
rational, 158; root of an, 46; 
transposition in an, 46. 

Equations, dependent, 91; incon¬ 
sistent, 91; independent, 79; 
simultaneous, 79; system of, 79. 

Exponent, 4; fractional, 172; 
negative, 174; zero, 174. 

Exponents, law of division of, 11; 
law of multiplication of, 8; 
laws of, 170, 171. 

Expression, algebraic, 3; mixed, 
42. 

Factor, 4; to, 24; theorem, 23,164. 

Factors, prime, 24. 

Formula, 61. 

Fractions, 29; adding, 36; clear¬ 
ing of, 49, 50; complex, 40; divi¬ 
sion of, 30; fundamental prin- 


* The numbers refer to pages. 
250 



INDEX 


251 


ciples of, 30; multiplication of, 
30; reducing, 30; subtracting, 
36. 

Function, 61; first degree, 65; 
second degree, 116. 

Functional relationship, 59. 

Geometric, mean, 206; progres¬ 
sion, 204. 

Graph of an equation with two 
variables, 145. 

Graphical solution of an equation 
with one variable, 118. 

Grouping, symbols of, 7; factoring 
by, 21. 

Horizontal axis, 64. 

Hyperbola, 149, 150. 

Identity, 46. 

Imaginary, numbers, 133; roots, 
132, 139; unit, 133. 

Inconsistent equations, 91. 

Independent equations, 79. 

Index, 177. 

Infinite geometric progression, 208. 

Irrational number, 137. 

Left side of an equation, 46. 

Like terms, 5. 

Logarithm, 184, 185. 

Mantissa, 186. 

Means, arithmetic, 197; geometric, 
206. 

Mixed expression, 42. 

Monomial, 4; addition of, 5; divi¬ 
sion of, 12; multiplication of, 8. 

Negative exponent, 174. 

Negative numbers, 2; addition of, 
2; division of, 2; multiplication 
of, 3; subtraction of, 2. 

Number, complex, 133; imaginary, 
133; negative, 2; positive, 2; 


prime, 24; rational, 137; real, 
133. 

Numerator, 30. 

Numerical coefficient, 4; identitv, 
46. 

Numerical value, 3. 

Ordinate, 64. 

Origin, 64. 

Parabola, 145. 

Parentheses, 7; inclosing terms in, 
7; removing, 7. 

Polynomial, 5. 

Polynomials, addition of, 5; divi¬ 
sion of, 12; factoring, 22; mul¬ 
tiplication of, 9; square root of, 
101; subtraction of, 5. 

Positive, number, 2. 

Power, 4. 

Powers, ascending, 9; descending, 9. 

Prime number, 24. 

Progression, arithmetic, 197; ge¬ 
ometric, 204. 

Quadratic equation, 116; com¬ 
plete, 116; forming an, 141; 
graph of an, 116, 118; having 
two unknowns, 145; imaginary 
roots of a, 132, 139; solution of, 
by completing the square, 122; 
by factoring, 120; by formula, 
126. 

Quadratic surd, 105. 

Radical, equation, 158; index of, 
177; sign, 100. 

Radicals, similar, 165. 

Radicand, 100. 

Ratio, of a geometric progression, 
204. 

Rational number, 137; polynomial, 
5. 

Rationalizing the denominator, 
109. 



252 


INDEX 


Reciprocal, 86. 

Remainder theorem, 165. 

Right side of an equation, 46. 
Root, cube, 172; of an equation, 
46; principal, 172; square, 100. 
Roots, imaginary, 132; product of, 
in a quadratic, 127; sum of, in a 
quadratic, 127. 

Rounding off numbers, 68. 

Sequence of terms, 196. 

Signs, change of, in an equation, 
48; law of, in addition, 2; in a 
fraction, 32; in division, 3; in 
multiplication, 3. 

Simultaneous equations, 79. 

Sine, 226. 

Solution of simultaneous equation, 
79, 80, 82; by determinants, 91. 
Square root, approximate, 103; by 
inspection, 100; on a monomial, 
100; of a number, 100; of a 
polynomial, 101; of a fraction, 

104. 

Surd, conjugate, 109; quadratic, 

105. 


Surds, addition of, 105; division 
of, 180; multiplication of, 181. 

Symbols of grouping, 7. 

Synthetic division, 166. 

System of equations, 75, 151. 

Table of square roots, 104; of 
logarithms, 188 

Tangent, 222. 

Term, 4. 

Terms, like, 5; unlike, 5. 

Theorem, binomial, 214; factor, 
164; remainder, 165. 

Transposition, 48. 

Trinomial, 5; perfect square, 122. 

Unit, imaginary, 133. 

Unlike terms, 5. 

Variable, 61; dependent, 61; in¬ 
dependent, 61. 

Variables, 61. 

Variation, 231. 

Varies, directly, 233; inversely, 
235; jointly, 234. 

Vertical axis, 64. 



Cnri^t(^ri^>^i;kt^.|Pki;^l^l^C4COCOCOCOWMMeOCOt9tOb9t9tOb9fc9tOtOMMMMMMMMMMI-» 

0(Ooo'>ao)c;i»(^cotdt-^oooo«ao>eni^cofcdi-^ou9oo<ia)cnH^cotoi-kOtooo>40)c;iif^&oboi-AO(00o«<iO)cni^ootOM 


TABLE I. Certain Powers AND Roots 


Sqs. 

Sq. 

Roots 

Cubes 

Cube 

Root.s 

No. 

Sqs. 

Sq. 

Roots 

Cubes 

Cube 

Root.s 

1 

1.000 

1 

1.000 

51 

2,601 

7.141 

132,651 

3.708 

4 

1.414 

8 

1.259 

52 

2,704 

7.211 

140,608 

3.732 

9 

1.732 

27 

1.442 

53 

2,809 

7.280 

148,877 

3.756 

16 

2.000 

64 

1.587 

54 

2,916 

7.348 

157,464 

3.779 

25 

2.236 

125 

1.709 

55 

3,025 

7.416 

166,375 

3.802 

36 

2.449 

216 

1.817 

56 

3,136 

7.483 

175,616 

3.825 

49 

2.645 

343 

1.912 

57 

3,249 

7.549 

185,193 

3.848 

64 

2.828 

512 

2.000 

58 

3,364 

7.615 

195,112 

3.870 

81 

3.000 

729 

2.080 

59 

3,481 

7.681 

205,379 

3.893 

100 

3.162 

1,000 

2.154 

60 

3,600 

7.745 

216,000 

3.914 

121 

3.316 

1,331 

2.223 

61 

3,721 

7.810 

226,981 

3.936 

144 

3.464 

1,728 

2.289 

62 

3,844 

7.874 

238,328 

3.957 

169 

3.605 

2,197 

2.351 

63 

3,969 

7.937 

250,047 

3.979 

196 

3.741 

2,744 

2.410 

64 

4,096 

8.000 

262,144 

4.000 

225 

3.872 

3,375 

2.466 

65 

4,225 

8.062 

274,625 

4.020 

256 

4.000 

4,096 

2.519 

66 

4,356 

8.124 

287,496 

4.041 

289 

4.123 

4,913 

2.571 

67 

4,489 

8.185 

300,763 

4.061 

324 

4.242 

5,832 

2.620 

68 

4,624 

8.246 

314,432 

4.081 

361 

4.358 

6,859 

2.668 

69 

4,761 

8.306 

328,509 

4.101 

400 

4.472 

8,000 

2.714 

70 

4,900 

8.366 

343,000 

4.121 

441 

4.582 

9,261 

2.758 

71 

5,041 

8.426 

357,911 

4.140 

484 

4.690 

10,648 

2.802 

72 

5,184 

8.485 

373,248 

4.160 

529 

4.795 

12,167 

2.843 

73 

5,329 

8.544 

389,017 

4.179 

576 

4.898 

13,824 

2.884 

74 

5,476 

8.602 

405,224 

4.198 

625 

5.000 

15,625 

2.924 

75 

5,625 

8.660 

421,875 

4.217 

676 

5.099 

17,576 

2.962 

76 

5,776 

8.717 

438,976 

4.235 

729 

5.196 

19,683 

3.000 

77 

5,929 

8.774 

456,533 

4.254 

784 

5.291 

21,952 

3.036 

78 

6,084 

8.831 

474,552 

4.272 

841 

5.385 

24,389 

3.072 

79 

6,241 

8.888 

493,039 

4.290 

900 

5.477 

27,000 

3.107 

80 

6,400 

8.944 

512,000 

4.308 

961 

5.567 

29,791 

3.141 

81 

6,561 

9.000 

531,441 

4.326 

1,024 

5.656 

32,768 

3.174 

82 

6,724 

9.055 

551,368 

4.344 

1,089 

5.744 

35,937 

3.207 

83 

6,889 

9.110 

571,787 

4.362 

1,156 

5.830 

39,304 

3.239 

84 

7,056 

9.165 

592,704 

4.379 

1,225 

5.916 

42,875 

3.271 

85 

7,225 

9.219 

614,125 

4.396 

1,296 

6.000 

46,656 

3.301 

86 

7,396 

9.273 

636,056 

4.414 

1,369 

6.082 

50,653 

3.332 

87 

7,569 

9.327 

658,503 

4.431 

1,444 

6.164 

54,872 

3.361 

88 

7,744 

9.380 

681,472 

4.447 

1,521 

6.245 

59,319 

3.391 

89 

7,921 

9.433 

704,969 

4.464 

1,600 

6.324 

64,000 

3.419 

90 

8,100 

9.486 

729,000 

4.481 

1,681 

6.403 

68,921 

3.448 

91 

8,281 

9.539 

753,571 

4.497 

1,764 

6.480 

74,088 

3.476 

92 

8,464 

9.591 

778,688 

4.514 

1,849 

6.557 

79,507 

3.503 

93 

8,649 

9.643 

804,357 

4.530 

1,936 

6.633 

85,184 

3.530 

94 

8,836 

9.695 

830,584 

4.546 

2,025 

6.708 

- 91,125 

3.556 

95 

9,025 

9.746 

857,375 

4.562 

2,116 

6.782 

97,336 

3.583 

96 

9,216 

9.797 

884,736 

4.578 

2,209 

6.855 

103,823 

3.608 

97 

9,409 

9.848 

912,673 

4.594 

2,304 

6.928 

110,592 

3.634 

98 

9,604 

9.899 

941,192 

4.610 

2,401 

7.000 

117,649 

3.659 

99 

9,801 

9.949 

970,299 

4.626 

2,500 

7.071 

125,000 

3.684 

100 

10,000 

10.000 

1,000,000 

4.641 


253 






















TABLE 11. Logarithms of Numbers 


Uo. 

0 

1 

2 

3 

4 

6 

6 

7 

8 

9 

10 

oooo 

0043 

0086 

0128 

0170 

0212 

0253 

0294 

0334 

0374 

IX 

0414 

0453 

0492 

0531 

0569 

0607 

0645 

0682 

0719 

0755 

12 

0792 

0828 

0864 

0899 

0934 

0969 

1004 

1038 

1072 

1106 

13 

1139 

1173 

1206 

1239 

1271 

1303 

1335 

1367 

1399 

1430 

14 

1461 

1492 

1523 

J553 

1584 

1614 

1644 

1673 

1703 

1732 

15 

1761 

1790 

1818 

1847 

1875 

1903 

1931 

1959 

1987 

2014 

i6 

2041 

2068 

2095 

2122 

2148 

2175 

2201 

2227 

2253 

2279 

17 

2304 

2330 

2355 

2380 

2405 

2430 

2455 

2480 

2504 

2529 

i8 

2 S 53 

2577 

•2601 

2625 

2648 

2672 

2695 

2718 

2742 

2765 

19 

2788 

2810 

2833 

2856 

2878 

2900 

2923 

2945 

2967 

2989 

20 

3010 

3032 

3054 

3075 

3096 

3118 

3139 

3160 

3181 

3201 

21 

3222 

3243 

3263 

3284 

3304 

3324 

3345 

3365 

3385 

3404 

22 

3424 

3444 

3464 

3483 

3502 

3522 

3541 

3560 

3579 

3598 

23 

3617 

3636 

3^55 

3674 

3692 

37 H 

3729 

3747 

3766 

3784 

24 

3802 

3820 

3838 

3856 

3874 

3892 

3909 

3927 

3945 

3962 

25 

3979 

3997 

4014 

4031 

4048 

4065 

4082 

4099 

4116 

4133 

26 

4150 

4166 

4183 

4200 

4216 

4232 

4249 

4265 

4281 

4298 

27 

4314 

4330 

4346 

4362 

4378 

4393 

4409 

4425 

4440 

4456 

28 

4472 

4487 

4502 

4 Si 8 

4533 

4548 

4564 

4579 

4594 

4609 

29 

4624 

4639 

4654 

4669 

4683 

4698 

4713 

4728 

4742 

4757 

30 

4771 

4786 

4800 

4814 

4829 

4843 

4857 

4871 

4886 

4900 

31 

4914 

4928 

4942 

4955 

4969 

4983 

4997 

50U 

5024 

5038 

32 

5051 

5065 

5079 

5092 

5105 

5119 

5132 

5145 

5159 

5172 

33 

5185 

5198 

5211 

5224 

5237 

5250 

5263 

5276 

5289 

5302 

34 

5315 

5328 

5340 

5353 

5366 

5378 

5391 

5403 

5416 

5428 

35 

5441 

5453 

5465 

5478 

5490 

5502 

5514 

5527 

5539 

5551 

36 

5563 

5575 

5587 

5599 

5611 

5623 

5635 

5647 

5658 

5670 

37 

5682 

5694 

5705 

5717 

5729 

5740 

5752 

5763 

5775 

5786 

38 

5798 

5809 

5821 

5832 

5843 

5855 

5866 

5877 

5888 

5899 

39 

5911 

5922 

5933 

5944 

5955 

5966 

5977 

5988 

5999 

6010 

40 

6021 

6031 

6042 

6053 

6064 

6075 

6085 

6096 

6107 

6117 

41 

6128 

6138 

6149 

6160 

6170 

6180 

6191 

6201 

6212 

6222 

42 

6232 

6243 

6253 

6263 

6274 

6284 

6294 

6304 

6314 

6325 

43 

6335 

6345 

6355 

6365 

6375 

6385 

6395 

6405 

6415 

6425 

44 

6435 

6444 

6454 

6464 

6474 

6484 

6493 

6503 

6513 

6522 

45 

6532 

6542 

6551 

6561 

6571 

6580 

6590 

6599 

6609 

6618 

46 

6628 

6637 

6646 

6656 

6665 

6675 

6684 

6693 

6702 

6712 

47 

6721 

6730 

6739 

6749 

6758 

6767 

6776 

6785 

6794 

6803 

48 

6812 

6821 

6830 

6839 

6848 

6857 

6866 

6875 

6884 

6893 

49 

6902 

6911 

6920 

6928 

6937 

6946 

6955 

6964 

6972 

6981 

50 

6990 

6998 

7007 

7016 

7024 

7033 

7042 

7050 

7059 

7067 

51 

7076 

7084 

7093 

7101 

7110 

7118 

7126 

7135 

7143 

7152 

52 

7160 

7168 

7177 

7185 

7193 

7202 

7210 

7218 

7226 

7235 

53 

7243 

7251 

7259 

7267 

7275 

7284 

7292 

7300 

7308 

7316 

54 

7324 

7332 

7340 

7348 

7356 

7364 

7372 

7380 

7388 

7396 

No. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

0 


254 





















TABLE II (Continued) 


No. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

55 

7404 

7412 

7419 

7427 

7435 

7443 

7451 

7459 

7466 

7474 

56 

7482 

7490 

7497 

7505 

7513 

7520 

7528 

7536 

7543 

755 ^ 

57 

7559 

7566 

7574 

7582 

7589 

7597 

7604 

7612 

7619 

7627 

58 

7634 

7642 

7649 

7657 

7664 

7672 

7679 

7686 

7694 

7701 

59 

7709 

7716 

7723 

7731 

7738 

7745 

7752 

7760 

7767 

7774 

60 

7782 

7789 

7796 

7803 

7810 

7818 

7825 

7832 

7839 

7846 

61 

7853 

7860 

7868 

7875 

7882 

7889 

7896 

7903 

7910 

7917 

62 

7924 

7931 

7938 

7945 

7952 

7959 

7966 

7973 

7980 

7987 

63 

7993 

8000 

8007 

8014 

802 1 

8028 

8035 

8041 

8048 

8055 

64 

8062 

8069 

8075 

8082 

8089 

8096 

8102 

8109 

8116 

8122 

65 

8129 

8136 

8142 

8149 

8156 

8162 

8169 

8176 

8182 

8189 

66 

8195 

8202 

8209 

8215 

8222 

8228 

8235 

8241 

8248 

8254 

67 

8261 

8267 

8274 

8280 

8287 

8293 

8299 

8306 

8312 

8319 

68 

8325 

8331 

8338 

8344 

8351 

8357 

8363 

8370 

8376 

8382 

69 

8388 

8395 

8401 

8407 

8414 

8420 

8426 

8432 

8439 

8445 

70 

8451 

8457 

8463 

8470 

8476 

8482 

8488 

8494 

8500 

8506 

71 

«Si 3 

8519 

8525 

8531 

8537 

8543 

8549 

8555 

8561 


72 


8 s 79 

8585 

8 S 9 I 

8597 

8603 

8609 

8615 

8621 

8627 

'n 

8633 

8639 

8645 

86s I 

8657 

8663 

8669 

8675 

8681 

8686 

74 

8692 

8698 

8704 

8710 

8716 

8722 

8727 

8733 

8739 

8745 

75 

8751 

8756 

8762 

8768 

8774 

8779 

8785 

8791 

8797 

8802 

76 

8808 

8814 

8820 

8825 

8831 

8837 

8842 

8848 

8854 

8859 

77 

8865 

8871 

8876 

8882 

8887 

8893 

8899 

8904 

8910 

8915 

78 

8921 

8927 

8932 

8938 

8943 

8949 

8954 

8960 

8965 

8971 

79 

8976 

8982 

8987 

8993 

8998 

9 cx )4 

9009 

9015 

9020 

9025 

80 

9031 

9036 

9042 

9047 

9053 

9058 

9063 

9069 

9074 

9079 

81 

9085 

9090 

9096 

9101 

9106 

9112 

9117 

9122 

9128 

9133 

82 

9138 

9143 

9149 

9154 

9159 

9165 

9170 

9175 

9180 

9186 

83 

9191 

9196 

9201 

9206 

9212 

9217 

9222 

9227 

9232 

9238 

84 

9243 

9248 

9253 

9258 

9263 

9269 

9274 

9279 

9284 

9289 

85 

9294 

9299 

9304 

9309 

9315 

9320 

9325 

9330 

9335 

9340 

86 

9345 

9350 

9355 

9360 

9365 

9370 

9375 

9380 

9385 

9390 

87 

9395 

9400 

9405 

9410 

9415 

9420 

9425 

9430 

9435 

9440 

88 

9445 

9450 

9455 

9460 

9465 

9469 

9474 

9479 

9484 

9489 

89 

9494 

9499 

9504 

9509 

9513 

951S 

9523 

9528 

9533 

9538 

90 

9542 

9547 

9552 

9557 

9562 

9566 

9571 

9576 

9581 

9586 

91 

9590 

9595 

9600 

9605 

9609 

9614 

9619 

9624 

9628 

9633 

92 

9638 

9643 

9647 

9652 

9657 

9661 

9666 

9671 

9675 

9680 

93 

9685 

9689 

9694 

9699 

9703 

9708 

9713 

9717 

9722 

9727 

94 

9731 

9736 

9741 

9745 

9750 

9754 

9759 

9763 

9768 

9773 

95 

9777 

9782 

9786 

9791 

9795 

9800 

9805 

9809 

9814 

9818 

96 

9823 

9827 

9832 

9836 

9841 

9845 

9850 

9854 

9859 

9863 

97 

9868 

9872 

9877 

9881 

9886 

9890 

9894 

9899 

9903 

9908 

98 

9912 

9917 

9921 

9926 

9930 

9934 

9939 

9943 

9948 

9952 

99 

9956 

9961 

9965 

9969 

9974 

9978 

9983 

9987 

9991 

9996 

Ho. 

0 

1 

2 

3 

4 

6 

6 

7 

8 

9 


255 



















THE LAW OF CO-FUNCTIONS IN TRIGONOMETRY 


1. Recall the following definitions: 

sin A = • cos A = —— * tan A = 

^ AB ’ ^ AB ’ ^ ^ AC 

2. Observe the following new definition : 

cotangent A = cot A = 



3. Observe that: 


sin B = 


AC. 

AB' 


cos B = 


BC . 
AB' 


tan B 


BC 


cot B = 


BC 

AC 


cos B = cos (90 — A), 
sin B = sin (90 — A), 
cot B = cot (90 — A). 

In general, any function of an angle equals the co¬ 
function of the complementary angle. 

5. This fact makes it possible to find the value of 
any function of an angle from 45° to 90° in a table of the 
values of the co-function of the angles from 0° to 45°. 

Thus cos 84° = sin 6°; and sin 82° 30' = cos 7° 30'. 


4. 


.-. sin A = ^ = 


cos A 


tan A = 


AB 

AC 

AB 

AC 


6. Table III is such a table. Observe its second 
column, headed by the abbreviation Sin. The numbers in 
this column are the sines of the angles opposite them at the 
left in column one. At the foot of column two is the abbre¬ 
viation Cos. This indicates that the numbers in column 
two are also the cosines of certain angles. They are the 
cosines of the angles opposite them at the right in column six. 

Thus: sin 6° = 0.1045 = cos 84°. 


7. To find the value of a function of an angle greater 
than 45°, find the angle in column six; find the desired 
function at the bottom of the page; then, above that func¬ 
tion, find the value which is opposite the given angle. 

Thus cos 67° 40' = .3800. (On page 259.) 

256 




TABLE III. Values of the Sines, Cosines, Tangents, 
AND Cotangents of Certain Angles 


Angle 

Sin 

Cos 

Tan 

Cot 

Angle 

0 ° 0' 

.0000 

1.0000 

.0000 


90° 0' 

10' 

.0029 

1.0000 

.0029 

343.77 

89° 50' 

20' 

.0058 

1.0000 

.0058 

171.89 

40' 

30' 

.0087 

1.0000 

.0087 

114.59 

30' 

40' 

.0116 

.9999 

.0116 

85.940 

20' 

50' 

.0145 

.9999 

.0145 

68.750 

89° 10' 

1° 0' 

.0175 

.9998 

.0175 

57.290 

89° 0' 

10' 

.0204 

.9998 

.0204 

49.104 

88° 50' 

20' 

.0233 

.9997 

.0233 

42.964 

40' 

30' 

.0262 

.9997 

.0262 

38.188 

30' 

40' 

.0291 

.9996 

.0291 

34.368 

20' 

50' 

.0320 

.9995 

.0320 

31.242 

88° 10' 

2° 0' 

.0349 

.9994 

.0349 

28.636 

88° 0' 

10' 

.0378 

.9993 

.0378 

26.432 

87° 50' 

20' 

.0407 

.9992 

.0407 

24.542 

40' 

30' 

.0436 

.9990 

.0437 

22.904 

30' 

40' 

.0465 

.9989 

.0466 

21.470 

20' 

50' 

.0494 

.9988 

.0495 

20.206 

87° 10' 

3° 0' 

.0523 , 

.9986 

.0524 

19.081 

87° 0' 

10' 

.0552 

.9985 

.0553 

18.075 

86° 50' 

20' 

.0581 

.9983 

.0582 

17.169 

40' 

30' 

.0610 

.9981 

.0612 

16.350 

30' 

40' 

.0640 

.9980 

.0641 

15.605 

20' 

50' 

.0669 

.9978 

.0670 

14.924 

86° 10' 

4° 0' 

.0698 

.9976 

.0699 

14.301 

86° 0' 

10' 

.0727 

i .9974 

.0729 

13.727 

85° 50' 

20' 

.0756 

1 .9971 

.0758 

13.197 

40' 

30' 

.0785 

.9969 

.0787 

12.706 

30' 

40' 

.0814 

.9967 

.0816 

12.251 

20' 

50' 

.0843 

.9964 

.0846 

11.826 

85° 10' 

5° 0' 

.0872 

.9962 

.0875 

11.430 

85° 0' 

10' 

.0901 

.9959 

.0904 

11.059 

84° 50' 

20' 

.0929 

.9957 

.0934 

10.712 

40' 

30' 

.0958 

.9954 

.0963 

10.385 

30' 

40' 

.0987 

.9951 

.0992 

10.078 

20' 

50' 

.1016 

.9948 

.1022 

9.7882 

84° 10' 

6° 0' 

! .1045 

.9945 

.1051 

9.5144 

84° 0' 

10' 

.1074 

.9942 

.1080 

9.2553 

83° 50' 

20' 

.1103 

.9939 

.1110 

9.0098 

40' 

30' 

.1132 

.9936 

.1139 

8.7769 

30' 

40' 

.1161 

.9932 

.1169 

8.5555 

20' 

50' 

.1190 

.9929 

.1198 

8.3450 

83° 10' 

70 0' 

.1219 

.9925 

.1228 

8.1443 

83° 0' 

10' 

.1248 

.9922 

.1257 

7.9530 

82° 50' 

20' 

.1276 

.9918 

.1287 

7.7704 

40' 

30' 

.1305 

.9914 

.1317 

7.5958 

30' 

40' 

.1334 

.9911 

.1346 

7.4287 

20' 

50' 

.1363 

.9907 

.1376 

7.2687 

82° 10' 

8° 0' 

.1392 

.9903 

.1405 

7.1154 

82° 0' 

10' 

.1421 

.9899 

.1435 

6.9682 

81° 50' 

20' 

.1449 

.9894 

.1465 

6.8269 

40' 

30' 

.1478 

.9890 

.1495 

6.6912 

30' 

40' 

.1507 

.9886 

.1524 

6.5606 

20' 

50' 

.1536 

.9881 

.1554 

6.4348 

81° 10' 

9° 0' 

.1564 

.9877 

.1584 

6.3138 

81° 0' 

Angle 

Cos 

Sin 

Cot 

Tan 

Angle 


257 



















TABLE III {Continued) 


Angle 

Sin 

Cos 

Tan 

1 Cot 

Angle 

9° 0' 

.1564 

.9877 

.1584 

_6.3138 

81° 0' 

10 ' 

.1593 

.9872 

.1614 

6.1970 

80° 50' 

20 ' 

.1622 

.9868 

.1644 

6.0844 

40' 

30' 

.1650 

.9863 

.1673 

5.9758 

30' 

40' 

.1679 

.9858 

.1703 

5.8708 

20 ' 

50' 

.1708 

.9853 

.1733 

5.7694 

80° 10' 

10 ° 0' 

.1736 

.9848 

.1763 

5.6713 

80° 0' 

10 ' 

.1765 

.9843 

.1793 

5.5764 

79° 50' 

20 ' 

.1794 

.9838 

.1823 

5.4845 

40' 

30' 

.1822 

.9833 

.1853 

5.3955 

30' 

40' 

.1851 

.9827 

.1883 

5.3093 

20 ' 

50' 

.1880 

.9822 

.1914 

5.2257 

79° 10' 

11 ° 0' 

.1908 

.9816 

.1944 

5.1446 

79° 0' 

10 ' 

.1937 

.9811 

.1974 

5.0658 

78° 50' 

20 ' 

.1965 

.9805 

.2004 

4.9894 

40' 

30' 

.1994 

.9799 

.2035 

4.9152 

30' 

40' 

.2022 

.9793 

.2065 

4.8430 

20 ' 

50' 

.2051 

.9787 

.2095 

4.7729 

78° 10' 

12 ° 0' 

.2079 

.9781 

.2126 

4.7046 

78° 0' 

10 ' 

.2108 

.9775 

.2156 

4.6382 

77° 50' 

20 ' 

.2136 

.9769 

.2186 

4.5736 

40' 

30' 

.2164 

.9763 

.2217 

4.5107 

30' 

40' 

.2193 

.9757 

.2247 

4.4494 

20 ' 

50' 

.2221 

.9750 

.2278 

4.3897 

77° 10' 

13° 0' 

.2250 

.9744 

.2309 

4.3315 

77° 0' 

10 ' 

.2278 

.9737 

.2339 

4.2747 

76° 50' 

20 ' 

.2306 

.9730 

.2370 

4.2193 

40' 

30' 

.2334 

.9724 

.2401 

4.1653 

30' 

40' 

.2363 

.9717 

.2432 

4.1126 

20 ' 

50' 

.2391 

.9710 

.2462 

4.0611 

76° 10' 

14° 0' 

.2419 

.9703 

.2493 

4.0108 

76° 0' 

10 ' 

.2447 

.9696 

.2524 

3.9617 

75° 50' 

20 ' 

.2476 

.9689 

.2555 

3.9136 

40' 

30' 

.2504 

.9681 

.2586 

3.8667 

30' 

40' 

.2532 

.9674 

.2617 

3.8208 

20 ' 

50' 

.2560 

.9667 

.2648 

3.7760 

75° 10' 

15° 0' 

.2588 

.9659 

.2679 

3.7321 

75° 0' 

10 ' 

.2616 

.9652 

.2711 

3.6891 

74° 50' 

20 ' 

.2644 

.9644 

.2742 

3.6470 

40' 

30' 

.2672 

.9636 

.2773 

3.6059 

30' 

40' 

.2700 

.9628 

.2805 

3.5656 

20 ' 

50' 

.2728 

.9621 

.2836 

3.5261 

74° 10' 

16° 0' 

.2756 

.9613 

.2867 

3.4874 

74° 0' 

10 ' 

.2784 

.9605 

.2899 

3.4495 

73° 50' 

20 ' 

.2812 

.9596 

.2931 

3.4124 

40' 

30' 

.2840 

.9588 

.2962 

3.3759 

30' 

40' 

.2868 

.9580 

.2994 

3.3402 

20 ' 

50' 

.2896 

.9572 

.3026 

3.3052 

73° 10' 

17° 0' 

.2924 

.9563 

.3057 

3.2709 

73° 0' 

10 ' 

.2952 

.9555 

.3089 

3.2371 

72° 50' 

20 ' 

.2979 

.9546 

.3121 

3.2041 

40' 

30' 

.3007 

.9537 

.3153 

3.1716 

30' 

40' 

.3035 

.9528 

.3185 

3.1.397 

20 ' 

50' 

.3062 

.9520 

.3217 

3.1084 

72° 10' 

18° 0' 

.3090 

.9511 

.3249 

3.0777 

72° 0' 

Angle 

Cos 

Sin 

Cot 

Tan 

Angle 


258 


























TABLE III {Continued) 


Angle 

Sin 

Cos 

Tan 

Cot 

Angle 

18° 0' 

.3090 

.9511 

.3249 

3.0777 

72° 0' 

10 ' 

.3118 

.9502 

.3281 

3.0475 

71° 50' 

20 ' 

.3145 

.9492 

.3314 

3.0178 

40' 

30' 

.3173 

.9483 

.3346 

2.9887 

30' 

40' 

.3201 

.9474 

.3378 

2.9600 

20 ' 

50' 

.3228 

.9465 

.3411 

2.9319 

71° 10' 

19° 0' 

.3256 

.9455 

.3443 

2.9042 

71° 0' 

10 ' 

.3283 

.9446 

.3476 

2.8770 

70° 50' 

20 ' 

.3311 

.9436 

.3508 

2.8502 

40' 

30' 

.3338 

.9426 

.3541 

2.8239 

30' 

40' 

.3365 

.9417 

.3574 

2.7980 

20 ' 

50' 

.3393 

.9407 

.3607 

2.7725 

70° 10' 

20 ° 0' 

.3420 

.9397 

.3640 

2.7475 

70° 0' 

10 ' 

.3448 

.9387 

.3673 

2.7228 

69° 50' 

20 ' 

.3475 

.9377 

.3706 

2.6985 

40' 

30' 

.3502 

.9367 

.3739 

2.6746 

30' 

40' 

.3529 

.9356 

.3772 

2.6511 

20 ' 

50' 

.3557 

.9346 

.3805 

2.6279 

69° 10' 

21 ° 0' 

.3584 

.9336 

.3839 

2.6051 

69° 0' 

10 ' 

.3611 

.9325 

.3872 

2.5826 

68 ° 50' 

20 ' 

.3638 

.9315 

.3906 

2.5605 

40' 

30' 

.3665 

.9304 

.3939 

2.5386 

30' 

40' 

.3692 

.9293 

.3973 

2.5172 

20 ' 

50' 

.3719 

.9283 

.4006 

2.4960 

68 ° 10' 

22 ° 0' 

.3746 

.9272 

.4040 

2.4751 

68 ° 0' 

10 ' 

.3773 

.9261 

.4074 

2.4545 

67° 50' 

20 ' 

.3800 

.9250 

.4108 

2.4342 

40' 

30' 

.3827 

.9239 

.4142 

2.4142 

30' 

40' 

.3854 

.9228 

.4176 

2.3945 

20 ' 

50' 

.3881 

.9216 

.4210 

2.3750 

67° 10' 

23° 0' 

.3907 

.9205 

.4245 

2.3559 

67° 0' 

10 ' 

.3934 

.9194 

.4279 

2.3369 

66 ° 50' 

20 ' 

.3961 

.9182 

.4314 

2.3183 

40' 

30' 

.3987 

.9171 

.4348 

2.2998 

30' 

40' 

.4014 

.9159 

.4383 

2.2817 

20 ' 

50' 

.4041 

.9147 

.4417 

2.2637 

66 ° 10' 

24° 0' 

.4067 

.9135 

.4452 

2.2460 

66 ° 0' 

10 ' 

.4094 

.9124 

.4487 

2.2286 

65° 50' 

20 ' 

.4120 

.9112 

.4522 

2.2113 

40' 

30' 

.4147 

.9100 

.4557 

2.1943 

30' 

40' 

.4173 

.9088 

.4592 

2.1775 

20 ' 

50' 

.4200 

.9075 

.4628 

2.1609 

65° 10' 

25° 0' 

.4226 

.9063 

.4663 

2.1445 

65° 0' 

10 ' 

.4253 

.9051 

.4699 

2.1283 

64° 50' 

20 ' 

.4279 

.9038 

.4734 

2.1123 

40' 

30' 

.4305 

.9026 

.4770 

2.0965 

30' 

40' 

.4331 

.9013 

.4806 

2.0809 

20 ' 

50' 

.4358 

.9001 

.4841 

2.0655 

64° 10' 

26° 0' 

.4384 

.8988 

.4877 

2.0503 

64° 0' 

10 ' 

.4410 

.8975 

.4913 

2.0353 

63° 50' 

20 ' 

.4436 

.8962 

.4950 

2.0204 

40' 

30' 

.4462 

.8949 

.4986 

2.0057 

30' 

40' 

.4488 

.8936 

.5022 

1.9912 

20 ' 

50' 

.4514 

.8923 

,5059 

1.9768 

63° 10' 

27° 0' 

.4540 

i .8910 

.5095 

1.9626 

63° 0' 

Angle 

‘Cos 

Sin 

Cot 

Tan 

Angle 


259 

























TABLE III {Continued) 


Angle 

Sin 

Cos 

Tan 

Cot 

Angle 

27° 0' 

.4540 

.8910 

.5095 

1.9626 

63° 0' 

10 ' 

.4566 

.8897 

.5132 

1.9486 

62° 50' 

20 ' 

.4592 

.8884 

.5169 

1.9347 

40' 

30' 

.4617 

.8870 

.5206 

1.9210 

30' 

40' 

.4643 

.8857 

.5243 

1.9074 

20 ' 

50' 

.4669 

.8843 

.5280 

1.8940 

62° 10' 

28° 0' 

.4695 

.8829 

.5317 

1.8807 

62° 0' 

10 ' 

.4720 

.8816 

.5354 

1.8676 

61° 50' 

20 ' 

.4746 

.8802 

.5392 

1.8546 

40' 

30' 

.4772 

.8788 

.5430 

1.8418 

30' 

40' 

.4797 

.8774 

.5467 

1.8291 

20 ' 

50' 

.4823 

.8760 

.5505 

1.8165 

61° 10' 

29° 0' 

.4848 

.8746 

.5543 

1.8040 

61° 0' 

10 ' 

.4874 

.8732 

.5581 

1.7917 

60° 50' 

20 ' 

.4899 

.8718 

.5619 

1.7796 

40' 

30' 

.4924 

.8704 

.5658 

1.7675 

30' 

40' 

.4950 

.8689 

.5696 

1.7556 

20 ' 

50' 

.4975 

.8675 

.5735 

1.7437 

60° 10' 

30° 0' 

.5000 

.8660 

.5774 

1.7321 

60° 0' 

10 ' 

.5025 

.8646 

.5812 

1.7205 

59° 50' 

20 ' 

.5050 

.8631 

.5851 

1.7090 

40' 

30' 

.5075 

.8616 

.5890 

1.6977 

30' 

40' 

.5100 

.8601 

.5930 

1.6864 

20 ' 

50' 

.5125 

.8587 

.5969 

1.6753 

59° 10' 

b 

o 

CO 

.5150 

.8572 

.6009 

1.6643 

59° 0' 

10 ' 

.5175 

.8557 

.6048 

1.6534 

58° 50' 

20 ' 

.5200 

.8542 

.6088 

1.6426 

40' 

30' 

.5225 

.8526 

.6128 

1.6319 

30' 

40' 

.5250 

.8511 

.6168 

1.6212 

20 ' 

50' 

.5275 

.8496 

.6208 

1.6107 

58° 10' 

32° 0' 

.5299 

.8480 

.6249 

1.6003 

58° 0' 

10 ' 

.5324 

.8465 

.6289 

1.5900 

57° 50' 

20 ' 

.5348 

.8450 

.6330 

1.5798 

40' 

30' 

.5373 

.8434 

.6371 

1.5697 

30' 

40' 

.5398 

.8418 

.6412 

1.5597 

20 ' 

50' 

.5422 

.8403 

.6453 

1.5497 

57° 10' 

33° 0' 

.5446 

.8387 

.6494 

1.5399 

57° 0' 

10 ' 

.5471 

.8371 

.6536 

1.5301 

56° 50' 

20 ' 

.5495 

.8355 

.6577 

1.5204 

40' 

30' 

.5519 

.8339 

.6619 

1.5108 

30' 

40' 

.5544 

.8323 

.6661 

1.5013 

20' 

50' 

.5568 

.8307 

.6703 

1.4919 

56° 10' 

34° 0' 

.5592 

.8290 

.6745 

1.4826 

56° 0' 

10 ' 

.5616 

.8274 

.6787 

1.4733 

55° 50' 

20 ' 

.5640 

.8258 

.6830 

1.4641 

40' 

30' 

.5664 

.8241 

.6873 

1.4550 

30' 

40' 

.5688 

.8225 

.6916 

1.4460 

20 ' 

50' 

.5712 

.8208 

.6959 

1.4370 

55° 10' 

35° 0' 

.5736 

.8192 

.7002 

1.4281 

55° 0' 

10 ' 

.5760 

.8175 

.7046 

1.4193 

54° 50' 

20 ' 

.5783 

.8158 

.7089 

1.4106 

40' 

30' 

.5807 

.8141 

.7133 

1.4019 

30' 

40' 

.5831 

.8124 

.7177 

1.3934 

20 ' 

50' 

.5854 

.8107 

.7221 

1.3848 

54° 10' 

36° 0' 

.5878 

.8090 

.7265 

1.3764 

54° 0' 

Angle 

Cos 

Sin 

Cot 

Tan 

Angle 


260 
















TABLE III {Continued) 


Angle 

Sin 

Cos 

Tan 

Cot 

Angle 

36° 0' 

.5878 

.8090 

.7265 

1.3764 

54° 0' 

10' 

.5901 

.8073 

.7310 

1.3680 

53° 50' 

20' 

.5925 

.8056 

.7355 

1.3597 

40' 

30' 

.5948 

.8039 

.7400 

1.3514 

30' 

40' 

.5972 

.8021 

.7445 

1.3432 

20' 

50' 

.5995 

.8004 

.7490 

1.3351 

53° 10' 

37° 0' 

.6018 

.7986 

.7536 

1.3270 

53° 0' 

10' 

.6041 

.7969 

.7581 

1.3190 

52° 50' 

20' 

.6065 

.7951 

.7627 

1.3111 

40' 

30' 

.6088 

.7934 

.7673 

1.3032 

30' 

40' 

.6111 

.7916 

.7720 

1.2954 

20' 

50' 

.6134 

.7898 

.7766 

1.2876 

52° 10' 

b 

00 

CO 

.6157 

.7880 

.7813 

1.2799 

52° 0' 

10' 

.6180 

.7862 

.7860 

1.2723 

51° 50' 

20' 

.6202 

.7844 

.7907 

1.2647 

40' 

30' 

.6225 

.7826 

.7954 

1.2572 

30' 

40' 

.6248 

.7808 

.8002 

1.2497 

20' 

50' 

.6271 

.7790 

.8050 

1.2423 

51° 10' 

39° 0' 

.6293 

.7771 

.8098 

1.2349 

51° 0' 

10' 

.6316 

.7753 

.8146 

1.2276 

50° 50' 

20' 

.6338 

.7735 

.8195 

1.2203 

40' 

30' 

.6361 

.7716 

.8243 

1.2131 

30' 

40' 

.6383 

.7698 

.8292 

1.2059 

20' 

50' 

.6406 

.7679 

.8342 

1.1988 

50° 10' 

o 

o 

O 

.6428 

.7660 

.8391 

1.1918 

50° 0' 

10' 

.6450 

.7642 

.8441 

1.1847 

49° 50' 

20' 

.6472 

.7623 

.8491 

1.1778 

40' 

30' 

.6494 

.7604 

.8541 

1.1708 

30' 

40' 

.6517 

.7585 

.8591 

1.1640 

20' 

50' 

.6539 

.7566 

.8642 

1.1571 

49° 10' 

0 

O 

.6561 

.7547 

.8693 

1.1504 

49° 0' 

10' 

.6583 

.7528 

.8744 

1.1436 

48° 50' 

20' 

.6604 

.7509 

.8796 

1.1369 

40' 

30' 

.6626 

.7490 

.8847 

1.1303 

30' 

40' 

.6648 

.7470 

.8899 

1.1237 

20' 

50' 

.6670 

.7451 

.8952 

1.1171 

48° 10' 

42° 0' 

.6691 

.7431 

.9004 

1.1106 

48° 0' 

10' 

.6713 

.7412 

.9057 

1.1041 

47° 50' 

20' 

.6734 

.7392 

.9110 

1.0977 

40' 

30' 

.6756 

.7373 

.9163 

1.0913 

30' 

40' 

.6777 

.7353 

.9217 

1.0850 

20' 

50' 

.6799 

.7333 

.9271 

1.0786 

47° 10' 

43° 0' 

.6820 

.7314 

.9325 

1.0724 

47° 0' 

10' 

.6841 

.7294 

.9380 

1.0661 

46° 50' 

20' 

.6862 

.7274 

.9435 

1.0599 

40' 

30' 

.6884 

.7254 

.9490 

1.0538 

30' 

40' 

.6905 

.7234 

.9545 

1.0477 

20' 

50' 

.6926 

.7214 

.9601 

1.0416 

46° 10' 

44° 0' 

.6947 

.7193 

.9657 

1.0355 

46° 0' 

10' 

.6967 

.7173 

.9713 

1.0295 

45° 50' 

20' 

.6988 

.7153 

.9770 

1.0235 

40' 

30' 

.7009 

.7133 

.9827 

1.0176 

30' 

40' 

.7030 

.7112 

.9884 

1.0117 

20' 

50' 

,7050 

.7092 

.9942 

1.0058 

45° 10' 

45° 0' 

.7071 

.7071 

1.0000 

1.0000 

45° 0' 

Angle 

1 Cos 

j Sin 

1 Cot 

1 

Tan 

Angle 


261 























TABLE IV. Logarithms op the Sines, Cosines, Tangents, 
AND Cotangents of Certain Angles (increased by 10) 


Angle 

Log sin 

Log cos 

Log tan 

Log cot 

Angle 

0® 0' 

10' 

7.4637 

10.0000 

7.4637 

12.5363 

90° 0' 

89® 50' 

20' 

.7648 

.0000 

.7648 

.2352 

40' 

30' 

.9408 

.0000 

.9409 

.0591 

30' 

40' 

8.0658 

.0000 

8.0658 

11.9342 

20' 

50' 

.1627 

.0000 

.1627 

.8373 

89® 10' 

1° 0' 

8.2419 

9.9999 

8.2419 

11.7581 

89° 0' 

10' 

.3088 

.9999 

.3089 

.6911 

88® 50' 

20' 

.3668 

.9999 

.3669 

.6331 

40' 

30' 

.4179 

.9999 

.4181 

.5819 

30' 

40' 

.4637 

.9998 

.4638 

.5362 

20' 

50' 

.5050 

.9998 

.5053 

.4947 

88® 10' 

2° 0' 

8.5428 

9.9997 

8.5431 

11.4569 

88® 0' 

10' 

.5776 

.9997 

.5779 

.4221 

87° 50' 

20' 

.6097 

.9996 

.6101 

.3899 

40' 

30' 

.6397 

.9996 

.6401 

.3599 

30' 

40' 

.6677 

.9995 

.6682 

.3318 

20' 

50' 

.6940 

.9995 

.6945 

.3055 

87® 10' 

3° 0' 

8.7188 

9.9994 

8.7194 

11.2806 

87° 0' 

10' 

.7423 

.9993 

.7429 

.2571 

86® 50' 

20' 

.7645 

.9993 

.7652 

.2348 

40' 

30' 

.7857 

.9992 

.7865 

.2135 

30' 

40' 

.8059 

.9991 

.8067 

.1933 

20' 

50' 

.8251 

.9990 

.8261 

.1739 

86® 10' 

4° 0' 

8.8436 

9.9989 

8.8446 

11.1554 

86° 0' 

10' 

.8613 

.9989 

.8624 

.1376 

85® 50' 

20' 

.8783 

.9988 

.8795 

.1205 

40' 

30' 

.8946 

.9987 

.8960 

.1040 

30' 

40' 

.9104 

.9986 

.9118 

.0882 

20' 

50' 

.9256 

.9985 

.9272 

.0728 

85® 10' 

5° 0' 

8.9403 

9.9983 

8.9420 

11.0580 

85® 0' 

10' 

.9545 

.9982 

.9563 

.0437 

84® 50' 

20' 

.9682 

.9981 

.9701 

.0299 

40' 

30' 

.9816 

.9980 

.9836 

.0164 

30' 

40' 

.9945 

.9979 

.9966 

.0034 

20' 

50' 

9.0070 

.9977 

9.0093 

10.9907 

84® 10' 

6® 0' 

9.0192 

9.9976 

9.0216 

.9784 

84° 0' 

10' 

.0311 

.9975 

.0336 

10.9664 

83® 50' 

20' 

.0426 

.9973 

.0453 

.9547 

40' 

30' 

.0539 

.9972 

.0567 

.9433 

30' 

40' 

.0648 

.9971 

.0678 

.9322 

20' 

50' 

.0755 

.9969 

.0786 

.9214 

83® 10' 

rjo 0' 

9.0859 

9.9968 

9.0891 

10.9109 

83° 0' 

10' 

.0961 

.9966 

.0995 

.9005 

82° 50' 

20' 

.1060 

.9964 

.1096 

.8904 

40' 

30' 

.1157 

.9963 

.1194 

.8806 

30' 

40' 

.1252 

.9961 

.1291 

.8709 

20' 

50' 

.1345 

.9959 

.1385 

.8615 

82® 10' 

8® 0' 

9.1436 

9.9958 

9.1478 

10.8522 

82® 0' 

10' 

.1525 

.9956 

.1569 

.8431 

81® 50' 

20' 

.1612 

.9954 

.1658 

.8342 

40' 

30' 

.1697 

.9952 

.1745 

.8255 

30' 

40' 

.1781 

.9950 

.1831 

.8169 

20' 

50' 

.1863 

.9948 

.1915 

.8085 

81® 10' 

9® 0' 

9.1943 

9.9946 

9.1997 

10.8003 

81® 0' 

Angle 

Log cos 

Log sin 

Log cot 

Log tan 

Angle 


262 















TABLE IV {Continued) 


Angle 

Log sin 

Log cos 

Log tan 

Log cot 

Angle 

9° 0' 

9.1943 

9.9946 

9.1997 

10.8003 

81° 0' 

10' 

.2022 

.9944 

.2078 

.7922 

80° 50' 

20' 

.2100 

.9942 

.2158 

.7842 

40' 

30' 

.2176 

.9940 

.2236 

.7764 

30' 

40' 

.2251 

.9938 

.2313 

.7687 

20' 

50' 

.2324 

.9936 

.2389 

.7611 

80° 10' 

b 

o 

O 

9.2397 

9.9934 

9.2463 

10.7537 

80° 0' 

10' 

.2468 

.9931 

.2536 

.7464 

79° 50' 

20' 

.2538 

.9929 

.2609 

.7391 

40' 

30' 

.2606 

.9927 

.2680 

.7320 

30' 

40' 

.2674 

.9924 

.2750 

.7250 

20' 

50' 

.2740 

.9922 

.2819 

.7181 

79° 10' 

11° 0' 

9.2806 

9.9919 

9.2887 

10.7113 

79° 0' 

10' 

.2870 

.9917 

.2953 

.7047 

78° 50' 

20' 

.2934 

.9914 

.3020 

.6980 

40' 

30' 

.2997 

.9912 

.3085 

.6915 

30' 

40' 

.3058 

.9909 

.3149 

.6851 

20' 

50' 

.3119 

.9907 

.3212 

.6788 

78° 10' 

12° 0' 

9.3179 

9.9904 

9.3275 

10.6725 

78° 0' 

10' 

.3238 

.9901 

.3336 

.6664 

77° 50' 

20' 

.3296 

.9899 

.3397 

.6603 

40' 

30' 

.3353 

.9896 

.3458 

.6542 

30' 

40' 

.3410 

.9893 

.3517 

.6483 

20' 

50' 

.3466 

.9890 

.3576 

.6424 

77° 10' 

13° 0' 

9.3521 

9.9887 

9.3634 

10.6366 

77° 0' 

10' 

.3575 

.9884 

.3691 

.6309 

76° 50' 

20' 

.3629 

.9881 

.3748 

.6252 

40' 

30' 

.3682 

.9878 

.3804 

.6196 

30' 

40' 

.3734 

.9875 

.3859 

.6141 

20' 

50' 

.3786 

.9872 

.3914 

.6086 

76° 10' 

o 

O 

9.3837 

9.9869 

9.3968 

10.6032 

76° 0' 

10' 

.3887 

.9866 

.4021 

.5979 

75° 50' 

20' 

.3937 

.9863 

.4074 

.5926 

40' 

30' 

.3986 

.9859 

.4127 

.5873 

30' 

40' 

.4035 

.9856 

.4178 

.5822 

20' 

50' 

.4083 

.9853 

.4230 

.5770 

75° 10' 

15° 0' 

9.4130 

9.9849 

9.4281 

10.5719 

75° 0' 

10' 

.4177 

.9846 

.4331 

.5669 

74° 50' 

20' 

.4223 

.9843 

.4381 

.5619 

40' 

30' 

.4269 

.9839 

.4430 

.5570 

30' 

40' 

.4314 

.9836 

.4479 

.5521 

20' 

50' 

.4359 

.9832 

.4527 

.5473 

74° 10' 

16° 0' 

9.4403 

9.9828 

9.4575 

10.5425 

74° 0' 

10' 

.4447 

.9825 

.4622 

.5378 

73° 50' 

20' 

.4491 

.9821 

.4669 

.5331 

40' 

30' 

.4533 

.9817 

.4716 

.5284 

30' 

40' 

.4576 

.9814 

.4762 

.5238 

20' 

50' 

.4618 

.9810 

.4808 

.5192 

73° 10' 

17° 0' 

9.4659 

9.9806 

9.4853 

10.5147 

73° 0' 

10' 

.4700 

.9802 

.4898 

.5102 

72° 50' 

20' 

.4741 

. 979 & 

.4943 

.5057 

40' 

30' 

.4781 

.9794 

.4987 

.5013 

30' 

40' 

.4821 

.9790 

.5031 

.4969 

20' 

50' 

.4861 

.9786 

.5075 

.4925 

72° 10' 

b 

o 

00 

9.4900 

9.9782 

9.5118 

10.4882 

72° 0' 

Angle ’ 

Log cos 

Log sin 

Log cot 

Log tan 

Angle 


263 



















TABLE IV {Continued) 


Angle 

Log sin 

Log cos 

Log tan 

Log cot 

Angle 

b 

o 

GO 

9.4900 

9.9782 

9.5118 

10.4882 

72° 0' 

10' 

.4939 

.9778 

.5161 

.4839 

71° 60' 

20' 

,4977 

.9774 

.5203 

.4797 

40' 

30' 

.5015 

.9770 

.5245 

.4755 

30' 

40' 

.5052 

.9765 

.5287 

.4713 

20' 

50' 

.5090 

.9761 

.5329 

.4671 

71° 10' 

19° 0' 

9.5126 

9.9757 

9.5370 

10.4630 

71° 0' 

10' 

.5163 

.9752 

.5411 

.4589 

70° 50' 

20' 

.5199 

.9748 

.5451 

.4549 

40' 

30' 

.5235 

.9743 

.5491 

.4509 

30' 

40' 

.5270 

.9739 

.5531 

.4469 

20' 

50' 

.5306 

.9734 

.5571 

.4429 

70° 10' 

20° 0' 

9.5341 

9.9730 

9.5611 

10.4389 

70° 0' 

10' 

.5375 

.9725 

.5650 

.4350 

69° 50' 

20' 

.5409 

.9721 

.5689 

.4311 

40' 

30' 

.5443 

.9716 

.5727 

.4273 

30' 

40' 

.5477 

.9711 

.5766 - 

.4234 

20' 

50' 

.5510 

.9706 

.5804 

.4196 

69° 10' 

b 

o 

9.5543 

9.9702 

9.5842 

10.4158 

69° 0' 

10' 

.5576 

.9697 

.5879 

.4121 

68° 50' 

20' 

.5609 

.9692 

.5917 

,4083 

40' 

30' 

.5641 

.9687 

.5954 

.4046 

30' 

40' 

.5673 

.9682 

.5991 

.4009 

20' 

50' 

.5704 

.9677 

.6028 

.3972 

68° 10' 

22° 0' 

9.5736 

9.9672 

9.6064 

10.3936 

68° 0' 

10' 

.5767 

.9667 

.6100 

.3900 

67° 50' 

20' 

.5798 

.9661 

.6136 

.3864 

40' 

30' 

.5828 

.9656 

.6172 

.3828 

30' 

40' 

.5859 

.9651 

.6208 

.3792 

20' 

50' 

.5889 

.9646 

.6243 

.3757 

67° 10' 

23° 0' 

9.5919 

9.9640 

9.6279 

10.3721 

67° 0' 

10' 

.5948 

.9635 

.6314 

.3686 

66° 50' 

20' 

.5978 

.9629 

.6348 

.3652 

40' 

30' 

.6007 

.9624 

.6383 

.3617 

30' 

40' 

.6036 

.9618 

.6417 

.3583 

20' 

50' 

.6065 

.9613 

.6452 

.3548 

66° 10' 

24° 0' 

9.6093 

9.9607 

9.6486 

10.3514 

66° 0' 

10' 

.6121 

.9602 

.6520 

.3480 

65° 50' 

20' 

.6149 

.9596 

.6553 

.3447 

40' 

30' 

.6177 

.9590 

.6587 

.3413 

30' 

40' 

.6205 

.9584 

.6620 

.3380 

20' 

50' 

.6232 

.9579 

.6654 

.3346 

65° 10' 

cn 

o 

O 

9.6259 

9.9573 

9.6687 

10.3313 

65° 0' 

10' 

.6286 

.9567 

.6720 

.3280 

64° 50' 

20' 

.6313 

.9561 

.6752 

.3248 

40' 

30' 

.6340 

.9555 

.6785 

.3215 

30' 

40' 

.6366 

.9549 

.6817 

.3183 

20' 

50' 

.6392 

.9543 

.6850 

.3150 

64° 10' 

26° 0' 

9.6418 

9.9537 

9.6882 

10.3118 

64° 0' 

10' 

.6444 

.9530 

.6914 

.3086 

63° 50' 

20' 

.6470 

.9524 

.6946 

.3054 

40' 

30' 

.6495 

.9518 

.6977 

.3023 

30' 

40' 

.6521 

.9512 

.7009 

.2991 

20' 

50' 

.6546 

.9505 

.7040 

.2960 

63° 10' 

27° 0' 

9.6570 

9.9499 

9.7072 

10.2928 

63° 0' 

Angle 

Log cos 

Log sin 

Log cot 

Log tan 

Angle 


264 

















TABLE IV {Continued) 


Angle 

Log sin 

Log cos 

Log tan 

Log cot 

Angle 

27 ° 0 ' 

9.6570 

9.9499 

9.7072 

10.2928 

63 ° 0 ' 

10 ' 

.6595 

.9492 

.7103 

.2897 

62 ° 60 ' 

20 ' 

.6620 

.9486 

.7134 

.2866 

40 ' 

30 ' 

.6644 

.9479 

.7165 

.2835 

30 ' 

40 ' 

.6668 

.9473 

.7196 

.2804 

20 ' 

50 ' 

.6692 

.9466 

.7226 

.2774 

62 ° 10 ' 

28 ° 0 ' 

9.6716 

' 9.9459 

9.7257 

10.2743 

62 ° 0 ' 

10 ' 

.6740 

.9453 

.7287 

.2713 

61 ° 50 ' 

20 ' 

.6763 

.9446 

.7317 

.2683 

40 ' 

30 ' 

.6787 

.9439 

.7348 

.2652 

30 ' 

40 ' 

.6810 

.9432 

.7378 

.2622 

20 ' 

50 ' 

.6833 

.9425 

.7408 

.2592 

61 ° 10 ' 

29 ° 0 ' 

9.6856 

9.9418 

9.7438 

10.2562 

61 ° 0 ' 

10 ' 

.6878 

.9411 

.7467 

.2533 

60 ° 50 ' 

20 ' 

.6901 

.9404 

.7497 

.2503 

40 ' 

30 ' 

.6923 

.9397 

.7526 

.2474 

30 ' 

40 ' 

.6946 

.9390 

.7556 

.2444 

20 ' 

50 ' 

.6968 

.9383 

.7585 

.2415 

60 ° 10 ' 

o 

o 

O 

CO 

9.6990 

9.9375 

9.7614 

10.2386 

60 ° 0 ' 

10 ' 

.7012 

.9368 

.7644 

.2356 

59 ° 50 ' 

20 ' 

.7033 

.9361 

.7673 

.2327 

40 ' 

30 ' 

.7055 

.9353 

.7701 

.2299 

30 ' 

40 ' 

.7076 

.9346 

.7730 

.2270 

20 ' 

50 ' 

.7097 

.9338 

.7759 

.2241 

59 ° 10 ' 

o 

o 

CO 

9.7118 

9.9331 

9.7788 

10.2212 

59 ° 0 ' 

10 ' 

.7139 

.9323 

.7816 

.2184 

58 ° 50 ' 

20 ' 

.7160 

.9315 

.7845 

.2155 

40 ' 

30 ' 

.7181 

.9308 

.7873 

.2127 

30 ' 

40 ' 

.7201 

.9300 

.7902 

.2098 

20 ' 

50 ' 

.7222 

.9292 

.7930 

.2070 

58 ° 10 ' 

32 ® 0 ' 

9.7242 

9.9284 

9.7958 

10.2042 

58 ° 0 ' 

10 ' 

.7262 

.9276 

.7986 

.2014 

57 ° 50 ' 

20 ' 

.7282 

.9268 

.8014 

.1986 

40 ' 

30 ' 

.7302 

.9260 

.8042 

.1958 

30 ' 

40 ' 

.7322 

.9252 

.8070 

.1930 

20 ' 

50 ' 

.7342 

.9244 

.8097 

.1903 

57 ° 10 ' 

33 ° 0 ' 

9.7361 

9.9236 

9.8125 

10.1875 

57 ° 0 ' 

10 ' 

.7380 

.9228 

.8153 

.1847 

56 ° 50 ' 

20 ' 

.7400 

.9219 

.8180 

.1820 

40 ' 

30 ' 

.7419 

.9211 

.8208 

.1792 

30 ' 

40 ' 

.7438 

.9203 

.8235 

.1765 

20 ' 

50 ' 

.7457 

.9194 

.8263 

.1737 

56 ° 10 ' 

34 ° 0 ' 

9.7476 

9.9186 

9.8290 

10.1710 

56 ° 0 ' 

10 ' 

.7494 

.9177 

.8317 

.1683 

55 ° 50 ' 

20 ' 

.7513 

.9169 

.8344 

.1656 

40 ' 

30 ' 

.7531 

.9160 

.8371 

.1629 

30 ' 

40 ' 

.7550 

.9151 

.8398 

.1602 

20 ' 

50 ' 

.7568 

.9142 

.8425 

.1575 

55 ° 10 ' 

35 ° 0 ' 

9.7586 

9.9134 

9.8452 

10.1548 

55 ° 0 ' 

10 ' 

.7604 

.9125 

.8479 

.1521 

54 ° 50 ' 

20 ' 

.7622 

.9116 

.8506 

.1494 

40 ' 

30 ' 

.7640 

.9107 

.8533 

.1467 

30 ' 

40 ' 

.7657 

.9098 

.8559 

.1441 

20 ' 

50 ' 

.7675 

.9089 

.8586 

.1414 

54 ° 10 ' 

36 ° 0 ' 

9.7692 

9.9080 

9.8613 

10.1387 

54 ° 0 ' 

Angle 

Log cos 

Log sin 

Log cot 

Log tan 

Angle 


265 



















TABLE IV {Continued) 


Angle 

Log sin 

Log cos 

Log tan 

Log cot 

Angle 

36° 0' 

9.7692 

9.9080 

9.8613 

10.1387 

en 

o 

O 

10' 

.7710 

.9070 

.8639 

.1361 

53° 50' 

20' 

.7727 

.9061 

.8666 

.1334 

40' 

30' 

.7744 

.9052 

.8692 

.1308 

30' 

40' 

.7761 

.9042 

.8718 

.1282 

20' 

50' 

.7778 

.9033 

.8745 

.1255 

53° 10' 

37° 0' 

9.7795 

9.9023 

9 . 8771 ' 

10.1229 

53° 0' 

10' 

.7811 

.9014 

.8797 

.1203 

52° 50' 

20' 

.7828 

.9004 

.8824 

.1176 

40' 

30' 

.7844 

.8995 

.8850 

.1150 

30' 

40' 

.7861 

.8985 

.8876 

.1124 

20' 

50' 

.7877 

.8975 

.8902 

.1098 

52° 10' 

00 

o 

O 

9.7893 

9.8965 

9.8928 

10.1072 

52° 0' 

10' 

.7910 

.8955 

.8954 

.1046 

51° 50' 

20' 

.7926 

.8945 

.8980 

.1020 

40' 

30' 

.7941 

.8935 

.9006 

.0994 

30' 

40' 

.7957 

.8925 

.9032 

.0968 

20' 

50' 

.7973 

.8915 

.9058 

.0942 

51° 10' 

39° 0' 

9.7989 

9.8905 

9.9084 

10.0916 

51° 0' 

10' 

.8004 

.8895 

.9110 

.0890 

50° 50' 

20' 

.8020 

.8884 

.9135 

.0865 

40' 

30' 

.8035 

.8874 

.9161 

.0839 

30' 

40' 

.8050 

.8864 

.9187 

.0813 

20' 

50' 

.8066 

.8853 

.9212 

.0788 

50° 10' 

o 

o 

O 

9.8081 

9.8843 

9.9238 

10.0762 

50° 0' 

10' 

.8096 

.8832 

.9264 

.0736 

49° 50' 

20' 

.8111 

.8821 

.9289 

.0711 

40' 

30' 

.8125 

.8810 

.9315 

.0885 

30' 

40' 

.8140 

.8800 

.9341 

.0659 

20' 

50' 

.8155 

.8789 

.9366 

.0634 

49° 10' 

b 

o 

9.8169 

9.8778 

9.9392 

10.0608 

49° 0' 

10' 

.8184 

.8767 

.9417 

.0583 

48° 50' 

20' 

.8198 

.8756 

.9443 

.0557 

40' 

30' 

.8213 

.8745 

.9468 

.0532 

30' 

40' 

.8227 

.8733 

.9494 

.0506 

20' 

50' 

.8241 

.8722 

.9519 

.0481 

48° 10' 

42° 0' 

9.8255 

9.8711 

9.9544 

10.0456 

48° 0' 

10' 

.8269 

.8699 

.9570 

.0430 

47° 50' 

20' 

.8283 

.8688 

.9595 

.0405 

40' 

30' 

.8297 

.8676 

.9621 

.0379 

30' 

40' 

.8311 

.8665 

.9646 

.0354 

20' 

50' 

.8324 

.8653 

.9671 

.0329 

47° 10' 

43° 0' 

9.8338 

9.8641 

9.9697 

10.0303 

47° 0' 

10' 

.8351 

.8629 

.9722 

.0278 

46° 50' 

20' 

.8365 

.8618 

.9747 

.0253 

40' 

30' 

.8378 

.8606 

.9773 

.0228 

30' 

40' 

.8391 

.8594 

.9798 

.0202 

20' 

50' 

.8405 

.8582 

.9823 

.0177 

46° 10' 

o 

O 

9.8418 

9.8569 

9.9848 

10.0152 

46° 0' 

10' 

.8431 

.8557 

.9874 

.0126 

45° 50' 

20' 

.8444 

.8545 

.9899 

.0101 

40' 

30' 

.8457 

.8532 

.9924 

.0076 

30' 

40' 

.8469 

.8520 

.9949 

.0051 

20' 

50' 

.8482 

.8507 

.9975 

.0025 

45° 10' 

45° 0' 

9.8495 

9.8495 

10.0000 

10.0000 

45° 0' 

Angle 

Log cos 

Log sin 

Log cot 

Log tan 

Angle 


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